Measurement of Angles
Group 'A' in BLE (1 Mark Each)
- समकोण त्रिभुजको सबैभन्दा लामो भुजा कुन हो ? What is the longest side of a right-angled triangle?
In a right-angled triangle, the longest side is the hypotenuse. It is the side opposite the right angle.
समकोण त्रिभुजमा सबैभन्दा लामो भुजा कर्ण (Hypotenuse) हो।
यो भुजा समकोण भएको कुनाको ठीक अगाडि (across from the right angle) हुन्छ।
- समकोण त्रिभुजका भुजाहरूको सम्बन्ध लेख्नुहोस्। Write the relation of the sides of a right-angled triangle?
The relation of the sides of a right-angled triangle is
\((\text{Perpendicular})^2 + (\text{Base})^2 = (\text{Hypotenuse})^2\)
\(p^2 + b^2=h^2 \)
where
p=one perpendicular side
b=other perpendicular side
h= Hypotenuse, i.e., the longest side opposite the right angle
समकोण त्रिभुजका भुजाहरूको सम्बन्ध निम्नानुसार हुन्छ
\((\text{Perpendicular})^2 + (\text{Base})^2 = (\text{Hypotenuse})^2\)
\(p^2 + b^2=h^2 \)
जहाँ
p = एक लम्ब भुजा (समकोणसँग मिल्ने)
b = अर्को लम्ब भुजा (समकोणसँग मिल्ने)
h = कर्ण (Hypotenuse), अर्थात् समकोणको विपरीत पर्ने सबैभन्दा लामो भुजा
- कोणलाई डिग्री एकाइमा नाप्ने पद्धतिको नाम के हो ? What is called the measurement system of angles in degree?
The measurement system of angles in degree is called "Sexagesimal system".There are three main systems used to measure angles: They are- Degree system or Sexagesimal system
- Grade system or Centisimal system
- Radian system or circular system
- कोणलाई ग्रेड एकाइमा नाप्ने पद्धतिलाई के भनिन्छ ? What is called the measurement system of angles in grade?
The measurement system of angles in grade is called "Centisimal system".There are three main systems used to measure angles: They are- Degree system or Sexagesimal system
- Grade system or Centisimal system
- Radian system or circular system
- \(90^\circ\) लाई ग्रेडमा बदल्नुहोस्। Convert \(90^\circ\) into grade.
We know that
1 right angle= 90 Degree =100 Grade
Therefore
\(90^\circ=100^g\) - \(63^\circ\) लाई ग्रेडमा बदल्नुहोस्। Convert \(63^\circ\) into grade.
We know that
1 right angle= 90 Degree =100 Grade
Therefore
\(63^\circ=63 \times \frac{100}{90} ^g=70^g\) - \(42.6^\circ\) लाई ग्रेडमा बदल्नुहोस्। Convert \(42.6^\circ\) into grade.
We know that
\(\pi^c= 180^o =200^g\)
Therefore
\(42.6^0=42.6 \times \frac{200}{180} ^g=47.33^g\) - \(45^\circ\) लाई रेडियनमा बदल्नुहोस्। Convert \(45^\circ\) into radian.
We know that
\(\pi^c= 180^o =200^g\)
Therefore
\(45^0=45 \times \frac{\pi}{180} ^c=\frac{\pi}{4}^c\) - \(\frac{3\pi}{5}\) लाई डिग्रीमा बदल्नुहोस्। Convert \(\frac{3\pi}{5}\) into degree.
We know that
\(\pi^c= 180^o =200^g\)
Therefore
\(\frac{3\pi}{5}^c=\frac{3\pi}{5} \times \frac{180}{\pi} ^o=108^o\) - \(80^g\) लाई रेडियनमा बदल्नुहोस्। Convert \(80^g\) into radian.
We know that
\(\pi^c= 180^o =200^g\)
Therefore
\(80^g=80 \times \frac{\pi}{200} ^c=\frac{2\pi}{5}^c\) - \(\frac{4\pi}{5}\) लाई ग्रेडमा बदल्नुहोस्। Convert \(\frac{4\pi}{5}\) into grade.
We know that
\(\pi^c= 180^o =200^g\)
Therefore
\(\frac{4\pi}{5}^c=\frac{4\pi}{5} \times \frac{200}{\pi} ^g=160^g\) - \(n\) ओटा भुजा भएको समबहुभुजको भित्री कोण डिग्रीमा पत्ता लगाउनुहोस्। Find the interior angle of a regular polygon of side \(n\) in degree.
The interior angle of a regular polygon of side \(n\) in degree is
\(\frac{(n-2)\times 180^0}{n}\)
Group 'B' in BLE (2 Marks Each)
- \(55^\circ 44' 40''\) लाई शंताशक सेकन्डमा बदल्नुहोस्। Convert \(55^\circ 44' 40''\) into centesimal second.
The conversion is
\(55^\circ 44' 40''\)
or\(55^\circ +44'+ 40''\)
or\(55 +\frac{44}{60}+ \frac{40}{60 \times 60}\) degree
or\(55.7444 \) degree
or\(55.7444 \times \frac{200}{180} \) grade
or\(61.9382 \) grade
or\(61.9382 \times 100 \times 100'' \)
or\(619382 '' \)
- \(55^g 44' 40''\) लाई शंताशक सेकन्डमा बदल्नुहोस्। Convert \(55^g 44' 40''\) into centesimal second.
The conversion is
\(55^g 44' 40''\)
or\(55^g +44'+ 40''\)
or\(55 \times 100 \times 100''+44 \times 100''+40'' \)
or\(554440'' \)
- \(27.5^0\) लाई ग्रेडमा बदल्नुहोस्। Convert \(27.5^0\) into grade.
We know that
\(\pi^c=180^o=200^g\)
Therefore,the conversion is
\(27.5^0\)
or\(27.5^0 \times \frac{200}{180}\) grade
or\(30.55^g \)
- \(10^\circ 30' 15''\) लाई डिग्रीमा बदल्नुहोस्। Reduce \(10^\circ 30' 15''\) into degree.
The conversion is
\(10^\circ 30' 15''\)
or\(10^\circ +30' +15''\)
or\(10+ \frac{30}{60}+ \frac{15}{60 \times 60} \) degree
or\(10.5042^\circ\)
- \(25^g 62' 50''\) लाई ग्रेडमा बदल्नुहोस्। Reduce \(25^g 62' 50''\) into grade.
The conversion is
\(25^g 62' 50''\)
or\(25^g+ 62'+ 50''\)
or\(25+ \frac{62}{100}+ \frac{50}{100 \times 100} \) grade
or\(25.6250^g\)
- \(12^\circ 22' 24''\) लाई षट्शांशक सेकेन्डमा बदल्नुहोस्। Convert \(12^\circ 22' 24''\) into sexagesimal second.
The conversion is
\(12^\circ 22' 24''\)
or\(12^\circ+ 22'+ 24''\)
or\(12 \times 60 \times 60'' +22 \times 60 '' +24''\)
or\(44544''\)
- \(36^\circ\) र \(24^\circ\) को अनुपात निकाल्नुहोस्। Find the ratio of \(36^\circ\) and \(24^\circ\).
The ratio of \(36^\circ\) and \(24^\circ\) is
\(\frac{36}{24}\)
or\(\frac{3}{2}\)
- दिएको सम्बन्धलाई ग्रेडमा रूपान्तरण गर्नुहोस्। Convert the following into grade: \(16^g 24' 36''\).
The conversion is
\(16^g 24' 36''\)
or\(16^g +24' +36''\)
or\(16^g +\frac{24}{100} +\frac{36}{100 \times 100}\) grade
or\(16.2436^g\)
- दिएको सम्बन्धलाई सेकेण्डमा रूपान्तरण गर्नुहोस्। Convert the following into second: \(16^g 24' 36''\).
The conversion is
\(16^g 24' 36''\)
or\(16^g +24' +36''\)
or\(16 \times 100 \times 100'' +24 \times 100 '' +36''\)
or\(162436''\)
- \(61.9444444^g\) लाई ग्रेड, मिनेट र सेकेन्डमा रूपान्तर गर्नुहोस्। Express \(61.9444444^g\) into grade, minute and second.
The conversion is
\(61.9444444^g\)
or\(16^g +0.9444444^g\)
or\(16^g +0.9444444 \times 100 '\)
or\(16^g +94.44444 '\)
or\(16^g +94'+0.44444'\)
or\(16^g +94'+0.44444 \times 100''\)
or\(16^g +94'+44.444 ''\)
or\(16^g 94' 44 ''\)
- \(61.9444444^g\) लाई षटदशांशक डिग्री, मिनेट र सेकेन्डमा रूपान्तर गर्नुहोस्। Express \(61.9444444^g\) into sexagesimal degree, minute and second.
The conversion is
\(61.9444444^g\)
or\(61.9444444 \times \frac{180}{200}\) degree
or\(55.7499\) degree
or\(55^o +0.7499^0\)
or\(55^o +0.7499 \times 60'\)
or\(16^o +44.994'\)
or\(16^o +44'+0.994'\)
or\(16^o +44'+0.994 \times 60''\)
or\(16^o +44'+59''\)
or\(16^o 44' 59''\)
- \(38.475^\circ\) लाई ग्रेड, मिनेट र सेकेन्डमा रूपान्तर गर्नुहोस्। (Express \(38.475^\circ\) into grade, minute and second.)
The conversion is
\(38.475^\circ\)
or\(38.475 \times \frac{200}{180}\) grade
or\(42.75\) degree
or\(55^o +0.7499^0\)
or\(55^o +0.7499 \times 60'\)
or\(16^o +44.994'\)
or\(16^o +44'+0.994'\)
or\(16^o +44'+0.994 \times 60''\)
or\(16^o +44'+59''\)
or\(16^o 44' 59''\)
- ग्रेडमा बदल्नुहोस् (Convert into grade): \(63^\circ 45'\)
The conversion is
\(63^\circ 45'\)
or\(63^\circ +45'\)
or\(63 +\frac{45}{60}\) degree
or\(63 .75\) degree
or\(63 .75 \times \frac{200}{180}\) grade
or\(70.8333\) grade
or\(70^g+0.8333^g\)
or\(70^g+0.8333\times 100 '\)
or\(70^g+83.33'\)
or\(70^g+83'+0.33'\)
or\(70^g+83'+0.33 \times 100''\)
or\(70^g+83'+33 ''\)
or\(70^g 83' 33 ''\)
- षट्दशांशक पद्धतिमा बदल्नुहोस् (Convert into sexagesimal system): \(90910^s\)
The conversion is
\(90910^s\)
or\(\frac{90910}{60}'\)
or\(1515'+10''\)
or\(\frac{1515}{60}^o+10''\)
or\(15^o+15'+ 10''\)
or\(15^o 15' 10''\)
- षट्दशांशक पद्धतिमा बदल्नुहोस् (Convert into sexagesimal system): \(90910^s\) of grade.
The conversion is
\(90910^s\) of grade
or\(\frac{90910}{10000}\) grade
or\(90.91\) grade
or\(90.91 \times \frac{180}{200}\) degree
or\(81.819\) degree
or\(81^0+0.819 \times 60 '\)
or\(81^0+49.14'\)
or\(81^0+49'+0.14\times 60''\)
or\(81^0+49'+8''\)
- शंताशक पद्धतिमा बदल्नुहोस् (Reduce into centesimal system): \(12.1525^\circ\)
The conversion is
\(12.1525^\circ\)
or\(12.1525\frac{200}{180}\) grade
or\(13.6127\) grade
or\(15^g 61' 27''\)
- \(27^g 62' 50''\) लाई षट्शांशक पद्धतिमा रूपान्तरण गर्नुहोस्। (Convert \(27^g 62' 50''\) into sexagesimal system.)
The conversion is
\(27^g 62' 50''\)
or\(27.6250\) grade
or\(27.6250 \times \frac{180}{200}\) degree
or\(24.8446\) degree
or\(24^0+0.8446 \times 60 '\)
or\(24^0+50.67'\)
or\(24^0+50'+0.67\times 60''\)
or\(24^0+50'+40''\)
- \(24^\circ 51' 45''\) लाई शंताशक पद्धतिमा रूपान्तरण गर्नुहोस्। (Convert \(24^\circ 51' 45''\) into centesimal system.)
The conversion is
\(24^\circ 51' 45''\)
\(24+ \frac{51}{60}+\frac{45}{60\times 60}\) degree
or\(24.8625\) degree
or\(24.8625\frac{200}{180}\) grade
or\(27.625\) grade
or\(27^g 62' 50''\)
- \(60^\circ 15' 30''\) लाई डिग्री र ग्रेडमा बदल्नुहोस्। (Express \(60^\circ 15' 30''\) into degree and grade.)
The conversion is
\(60^\circ 15' 30''\)
\(60+ \frac{15}{60}+\frac{30}{60\times 60}\) degree
or\(60.2583\) degree
Next, the grade conversion is
or\(60.2583\frac{200}{180}\) grade
or\(66.953\) grade
- \(27^g 62' 50''\) लाई ग्रेड र डिग्रीमा बदल्नुहोस्। (Convert \(27^g 62' 50''\) into grade and degree.)
The conversion is
\(27^g 62' 50''\)
\(27+ \frac{62}{100}+\frac{50}{100\times 100}\) grade
or\(27.6250\) grade
Next, the degree conversion is
or\(27.6250\frac{180}{200}\) degree
or\(24.8625\) degree
- दिइएको सम्बन्धलाई ग्रेड र डिग्रीमा बदल्नुहोस्। (Express the following relation into grade and degree): \(56^g 87' 50''\)
The conversion is
\(56^g 87' 50''\)
\(56+ \frac{87}{100}+\frac{50}{100\times 100}\) grade
or\(56.8750\) grade
Next, the degree conversion is
or\(56.8750\frac{180}{200}\) degree
or\(51.1875\) degree
- तलको सम्बन्धलाई डिग्री र ग्रेडमा बदल्नुहोस्। (Express the following relation into degree and grade): \(44^\circ 16' 5''\)
The conversion is
\(44^\circ 16' 5''\)
\(44+ \frac{16}{60}+\frac{5}{60\times 60}\) degree
or\(44.2681\) degree
Next, the grade conversion is
or\(44.2681\frac{200}{180}\) grade
or\(49.1867\) grade
- \(16^\circ\) र \(20^g\) लाई रेडियन नापमा व्यक्त गरी अनुपात पत्ता लगाउनुहोस्। (Express \(16^\circ\) and \(20^g\) in radian and also find the ratio.)
The conversion is
\(16^\circ=16\times \frac{\pi}{180} =\frac{4 \pi}{45}\) radian
\(20^g=20\times \frac{\pi}{200} =\frac{\pi}{10}\) radian
Next, the ratio is
or\(\frac{\frac{4 \pi}{45}}{\frac{\pi}{10}} \) or \(40:45\) or \(8:9\)
- \(\frac{\pi^c}{12}\) र \(18^\circ\) लाई ग्रेडमा व्यक्त गरी अनुपात पत्ता लगाउनुहोस्। (Express \(\frac{\pi^c}{12}\) and \(18^\circ\) in grade, and find the ratio.)
The conversion is
\(\frac{\pi^c}{12}=\frac{\pi}{12}\times \frac{200}{\pi} =\frac{50}{3}\) grade
\(18^o=18\times \frac{200}{180} =20\) grade
Next, the ratio is
or\(\frac{\frac{50}{3}}{20} \) or \(50:60\) or \(5:6\)
Group 'C' in BLE (4 Marks Each)
- समकोण त्रिभुजको एउटा कोण \(30^\circ\) भए, अर्को कोण डिग्रीमा पत्ता लगाउनुहोस्। (If one angle of a right-angled triangle is \(30^\circ\), find the other angle in degree.)
In a right angled triangle
\(\measuredangle A=90^o\)
\(\measuredangle B=30^o\)
\(\measuredangle C=x\)
We know that
\(90+30+x=180\)
or\(x=60^o\)
- त्रिभुजका दुई कोणहरू \(40^g\) र \(100^\circ\) भए, बाँकी कोण डिग्रीमा पत्ता लगाउनुहोस्। (If two angles of a triangle are \(40^g\) and \(100^\circ\), find the remaining angles in degree.)
In a triangle
\(\measuredangle A=100^o\)
\(\measuredangle B=40^g=40 \times \frac{9}{10}=36^o\)
\(\measuredangle C=x\)
We know that
\(100+36+x=180\)
or\(x=44^o\)
- यदि समकोणी त्रिभुजको एक कोण \(54^\circ\) भए, सबै कोणहरूलाई ग्रेडमा पत्ता लगाउनुहोस्। (If one angle of a right-angled triangle is \(54^\circ\), find all the angles in grade.)
In a triangle
\(\measuredangle A=90^0=100^o\)
\(\measuredangle B=54^0=54 \times \frac{10}{9}=60^g\)
\(\measuredangle C=x\)
We know that
\(100+60+x=200\)
or\(x=40^g\)
- यदि समकोणी त्रिभुजको एक कोण \(30^g\) भए, सबै कोणहरूलाई डिग्रीमा पत्ता लगाउनुहोस्। (If one angle of a right-angled triangle is \(30^g\), find all the angles in degree.)
In a triangle
\(\measuredangle A=90^0\)
\(\measuredangle B=30^g=30 \times \frac{9}{10}=27^o\)
\(\measuredangle C=x\)
We know that
\(90+27+x=180\)
or\(x=63^0\)
- समकोणी त्रिभुजको एउटा कोण \(\left(\frac{\pi}{18}\right)\) भए, अर्को कोण डिग्रीमा पत्ता लगाउनुहोस्। (If one angle of a right-angled triangle is \(\left(\frac{\pi}{18}\right)\) radian, find the another angle in degree.)
In a triangle
\(\measuredangle A=90^0\)
\(\measuredangle B=\frac{\pi}{18}^c=\frac{\pi}{18} \times \frac{180}{\pi}=10^o\)
\(\measuredangle C=x\)
We know that
\(90+10+x=180\)
or\(x=80^0\)
- यदि त्रिभुजका दुई कोणहरू \(63^\circ\) र \(72^\circ\) भए, सबै कोणहरू रेडियनमा पत्ता लगाउनुहोस्। (If two angles of a triangle are \(63^\circ\) and \(72^\circ\), find all the angles in radian.)
In a triangle
\(\measuredangle A=72^0\)
\(\measuredangle B=63^o\)
\(\measuredangle C=x\)
We know that
\(72+63+x=180\)
or\(x=45^0\)
Hence the angles are
\(\measuredangle A=72^0=72 \times \frac{\pi}{180}=\frac{8 \pi}{20}\) radian
\(\measuredangle B=63^0=63 \times \frac{\pi}{180}=\frac{7 \pi}{20}\) radian
\(\measuredangle C=45^0=45 \times \frac{\pi}{180}=\frac{5 \pi}{20}\) radian
- त्रिभुजका दुई कोणहरू \(2:7\) को अनुपातमा छन् र तेस्रो कोण \(80^\circ\) भए, सबै कोणहरू डिग्रीमा पत्ता लगाउनुहोस्। (Two angles of a triangle are in the ratio \(2:7\) and the third angle is \(80^\circ\), find all the angles in degree.)
In a triangle
\(\measuredangle A=84^0\)
\(\measuredangle B=x\)
\(\measuredangle C=3x\)
We know that
\(84+x+3x=180\)
or\(x=24^0\)
Hence the angles are
\(\measuredangle A=84^0\)
\(\measuredangle B=x=24^0\)
\(\measuredangle C=3x=3 \times 72^0\)
- त्रिभुजको एउटा कोण \(27^\circ\) र बाँकी कोणहरू \(8:9\) को अनुपातमा भए, सबै कोणहरू ग्रेडमा पत्ता लगाउनुहोस्। (One angle of a triangle is \(27^\circ\) and the remaining angles are in the ratio \(8:9\), find all the angles in grade.)
In a triangle
\(\measuredangle A=9x\)
\(\measuredangle B=8x\)
\(\measuredangle C=27^0=27 \times \frac{10}{9}=30^g\)
We know that
\(30+8x+9x=200\)
or\(x=10^g\)
Hence the angles are
\(\measuredangle A=9x=90^g\)
\(\measuredangle B=8x=80^g\)
\(\measuredangle C=30^g\)
- एउटा त्रिभुजका दुई कोणहरू \(1:2\) को अनुपातमा छन् र तेस्रो कोण \(60^\circ\) भए, सबै कोणहरू रेडियन नापमा पत्ता लगाउनुहोस्। (Two angles of a triangle are in the ratio \(1:2\) and the third angle is \(60^\circ\), find the angles in radian.)
In a triangle
\(\measuredangle A=60^o\)
\(\measuredangle B=2x\)
\(\measuredangle C=1x\)
We know that
\(60+2x+1x=180\)
or\(x=40^o\)
Hence the angles are
\(\measuredangle A=60\times \frac{\pi}{180}=\frac{\pi}{3}\) radian
\(\measuredangle B=2x=80^o=80\times \frac{\pi}{180}=\frac{4\pi}{9}\) radian
\(\measuredangle C=1x=40^o=40\times \frac{\pi}{180}=\frac{2\pi}{9}\) radian
- यदि त्रिभुजका दुई कोणहरू \(50^\circ\) र \(100^\circ\) भए, तेस्रो कोण रेडियनमा पत्ता लगाउनुहोस्। (If two angles of a triangle are \(50^\circ\) and \(100^\circ\), find the third angle in radian.)
In a triangle
\(\measuredangle A=100^o\)
\(\measuredangle B=50^o\)
\(\measuredangle C=x\)
We know that
\(100+50+x=180\)
or\(x=30^o\)
Hence the angle is
\(\measuredangle C=x=30^o=30\times \frac{\pi}{180}=\frac{\pi}{6}\) radian
- यदि त्रिभुजका दुई कोणहरू \(50^\circ\) र \(80^g\) भए, तेस्रो कोण रेडियनमा पत्ता लगाउनुहोस्। (If two angles of a triangle are \(50^\circ\) and \(80^g\), find the third angle in radian.)
In a triangle
\(\measuredangle A=50^o=50 \times \frac{\pi}{180}=\frac{5\pi}{18}\)
\(\measuredangle B=80^g \times \frac{\pi}{200}=\frac{2\pi}{5}\)
\(\measuredangle C=x\)
We know that
\(\frac{5\pi}{18}+\frac{2\pi}{5}+x=\pi\)
or\(x=\frac{29\pi}{90}\)
Hence the angle is
\(\measuredangle C=x=\frac{29\pi}{90}\) radian
- यदि त्रिभुजका दुई कोणहरू \(\left(\frac{4\pi}{9}\right)^c\) र \(40^\circ\) भए, तेस्रो कोण रेडियनमा पत्ता लगाउनुहोस्। (If two angles of a triangle are \(\left(\frac{4\pi}{9}\right)^c\) and \(40^\circ\), find the third angle in radian.)
In a triangle
\(\measuredangle A=\frac{4\pi}{9}\)
\(\measuredangle B=40^o=40 \times \frac{\pi}{180}=\frac{2\pi}{9}\)
\(\measuredangle C=x\)
We know that
\(\frac{4\pi}{9}+\frac{2\pi}{9}+x=\pi\)
or\(x=\frac{3\pi}{9}\)
Hence the angle is
\(\measuredangle C=x=\frac{3\pi}{9}\) radian
- कुनै त्रिभुजको एक कोण एक समकोणको \(\frac{2}{3}\) भाग छ। बाँकी दुई कोणहरू एकअर्कासँग \(18^\circ\) ले बढी छन् भने सबै कोणहरू डिग्रीमा पत्ता लगाउनुहोस्। (One angle of a triangle is \(\frac{2}{3}\) of a right-angle. If the greater of the other two exceeds the smaller by \(18^\circ\), find all the angles in degree.)
In a triangle
\(\measuredangle A=90 \times \frac{2}{3}=60^0\)
\(\measuredangle B=x\) degree
\(\measuredangle C=x+18\) degree
We know that
\(60+x+x+18=180\)
or\(x=51\)
Hence the angles are
\(\measuredangle B=x=51^0\)
\(\measuredangle C=x+18=59^0\)
- एउटा त्रिभुजका तीन कोणहरू \(1:2:3\) को अनुपातमा छन्। प्रत्येक कोणहरू डिग्रीमा पत्ता लगाउनुहोस्। (The three angles of a triangle are in the ratio \(1:2:3\), find the angles in degree.)
In a triangle
\(\measuredangle A=3x\)
\(\measuredangle B=2x\) degree
\(\measuredangle C=x\) degree
We know that
\(x+2x+3x=180\)
or\(x=30\)
Hence the angles are
\(\measuredangle A=3x=90^0\)
\(\measuredangle B=x=60^0\)
\(\measuredangle C=x=30^0\)
- यदि त्रिभुजका तीन कोणहरू \(2:3:5\) को अनुपातमा छन् भने प्रत्येक कोणको ग्रेडमा पत्ता लगाउनुहोस्। (If three angles of a triangle are in the ratio \(2:3:5\), find each of the angles in grade.)
In a triangle
\(\measuredangle A=5x\)
\(\measuredangle B=3x\) degree
\(\measuredangle C=2x\) degree
We know that
\(2x+3x+5x=200\)
or\(x=20\)
Hence the angles are
\(\measuredangle A=5x=100^g\)
\(\measuredangle B=3x=60^g\)
\(\measuredangle C=2x=40^g\)
Right Angled Triangle and Trigonometric Ratios
Group 'B' in BLE (2 Marks Each)
- पाइथागोरस साध्य प्रयोग गरी \(\sin^2\theta + \cos^2\theta = 1\) प्रमाणित गर्नुहोस्।
(Prove by Pythagoras theorem that: \(\sin^2\theta + \cos^2\theta = 1\))
Consider a right-angled triangle \(ABC\) with right angle at \(B\), in which \(\angle A = \theta\).
Then,
Opposite side to \(\theta\) = \(BC = p\) (perpendicular)
Adjacent side to \(\theta\) = \(AB = b\) (base)
Hypotenuse = \(AC = h\)
By definition of trigonometric ratios, we get
\(\sin \theta = \dfrac{p}{h}\)
\(\cos \theta = \dfrac{b}{h}\)
Now, by Pythagoras theorem, we have
\(p^2 + b^2 = h^2\)
Divide both sides by \(h^2\), we get
\(\dfrac{p^2}{h^2} + \dfrac{b^2}{h^2} = \dfrac{h^2}{h^2}\)
or\(\left( \dfrac{p}{h} \right)^2 + \left( \dfrac{b}{h} \right)^2 = 1\)
or\(\sin^2 \theta + \cos^2 \theta = 1\)
Hence, proved.
- पाइथागोरस साध्य प्रयोग गरी \(\sec^2\theta - \tan^2\theta = 1\) प्रमाणित गर्नुहोस्।
(Prove by Pythagoras theorem that: \(\sec^2\theta - \tan^2\theta = 1\))
Consider a right-angled triangle \(ABC\) with right angle at \(B\), in which \(\angle A = \theta\).
Then,
Opposite side to \(\theta\) = \(BC = p\) (perpendicular)
Adjacent side to \(\theta\) = \(AB = b\) (base)
Hypotenuse = \(AC = h\)
By definition of trigonometric ratios, we get
\(\tan \theta = \dfrac{p}{b}\)
\(\sec \theta = \dfrac{h}{b}\)
Now, by Pythagoras theorem, we have
\( h^2- p^2=b^2 \)
Divide both sides by \(b^2\), we get
\(\dfrac{h^2}{b^2} - \dfrac{p^2}{b^2} = \dfrac{b^2}{b^2}\)
or\(\left( \dfrac{h}{b} \right)^2 - \left( \dfrac{p}{b} \right)^2=1\)
or\(\sec^2 \theta - \tan^2 \theta = 1\)
Hence, proved.
- पाइथागोरस साध्य प्रयोग गरी \(\csc^2\theta - \cot^2\theta = 1\) प्रमाणित गर्नुहोस्।
(Prove by Pythagoras theorem that: \(\csc^2\theta - \cot^2\theta = 1\))
Consider a right-angled triangle \(ABC\) with right angle at \(B\), in which \(\angle A = \theta\).
Then,
Opposite side to \(\theta\) = \(BC = p\) (perpendicular)
Adjacent side to \(\theta\) = \(AB = b\) (base)
Hypotenuse = \(AC = h\)
By definition of trigonometric ratios, we get
\(\cot \theta = \dfrac{b}{p}\)
\(\csc \theta = \dfrac{h}{p}\)
Now, by Pythagoras theorem, we have
\(h^2 - b^2 = p^2\)
Divide both sides by \(p^2\), we get
\(\dfrac{h^2}{p^2} - \dfrac{b^2}{p^2} = \dfrac{p^2}{p^2}\)
or\(\left( \dfrac{h}{p} \right)^2 - \left( \dfrac{b}{p} \right)^2 = 1\)
or\(\csc^2 \theta - \cot^2 \theta = 1\)
Hence, proved.
- पाइथागोरस साध्य प्रयोग गरी \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) प्रमाणित गर्नुहोस्।
(Prove \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) by using Pythagoras theorem.)
Consider a right-angled triangle \(ABC\) with right angle at \(B\), in which \(\angle A = \theta\).
Then,
Opposite side to \(\theta\) = \(BC = p\)
Adjacent side to \(\theta\) = \(AB = b\)
Hypotenuse = \(AC = h\)
By definition of trigonometric ratios, we have
\(\sin \theta = \dfrac{p}{h}\)
\(\cos \theta = \dfrac{b}{h}\)
\(\tan \theta = \dfrac{p}{b}\)
Now,
\(\dfrac{\sin \theta}{\cos \theta} =\dfrac{\dfrac{p}{h}}{\dfrac{b}{h}}=\dfrac{p}{h} \cdot \dfrac{h}{b} =\dfrac{p}{b}=\tan \theta\)
Thus,
\(\tan \theta = \dfrac{\sin \theta}{\cos \theta}\)
Hence, proved
- दिएको समकोणी त्रिभुजबाट \(\sin\theta\), \(\cos\theta\) र \(\tan\theta\) को मान पत्ता लगाउनुहोस्।
(Find the values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\) from the given right-angled triangle.)
In the given right-angled triangle \(ABC\), the right angle is at \(B\).
Given: \(AB = 3\), \(BC = 4\), \(AC = 5\).
Let \(\angle A = \theta\).
Then,
- Side opposite to \(\theta\) = \(BC = 4\)
- Side adjacent to \(\theta\) = \(AB = 3\)
- Hypotenuse = \(AC = 5\)
Now, using definitions of trigonometric ratios:
\(\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{4}{5}\)
\(\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{3}{5}\)
\(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{4}{3}\)
Therefore,
\(\sin \theta = \dfrac{4}{5},\quad \cos \theta = \dfrac{3}{5},\quad \tan \theta = \dfrac{4}{3}\)
- यदि \(\sin\theta = \dfrac{1}{\sqrt{2}}\) भए, \(\csc\theta\) को मान निकाल्नुहोस्।
(If \(\sin\theta = \dfrac{1}{\sqrt{2}}\), find the value of \(\csc\theta\).)
We know that cosecant is the reciprocal of sine:
\(\csc \theta = \dfrac{1}{\sin \theta}\)
Given:
\(\sin \theta = \dfrac{1}{\sqrt{2}}\)
Therefore,
\(\csc \theta = \dfrac{1}{\dfrac{1}{\sqrt{2}}} = \sqrt{2}\)
Hence, the value of \(\csc \theta\) is \(\sqrt{2}\).
- यदि \(\cos\theta = \dfrac{3}{5}\) भए, \(\sin\theta\) को मान निकाल्नुहोस्।
(If \(\cos\theta = \dfrac{3}{5}\), find the value of \(\sin\theta\).)
We use the Pythagorean identity:
\(\sin^2 \theta + \cos^2 \theta = 1\)
Given:
\(\cos \theta = \dfrac{3}{5}\)
Then,
\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left( \dfrac{3}{5} \right)^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25}\)
Taking square root on both sides:
\(\sin \theta = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}\)
- यदि \(\sec\theta = \dfrac{13}{5}\) भए, \(\tan\theta\) को मान निकाल्नुहोस्।
(If \(\sec\theta = \dfrac{13}{5}\), find the value of \(\tan\theta\).)
We use the Pythagorean identity:
\(\sec^2 \theta - \tan^2 \theta = 1\)
Given:
\(\sec \theta = \dfrac{13}{5}\)
Then,
\(\tan^2 \theta = \sec^2 \theta - 1 = \left( \dfrac{13}{5} \right)^2 - 1 = \dfrac{169}{25} - \dfrac{25}{25} = \dfrac{144}{25}\)
Taking square root on both sides, we get
\(\tan \theta = \sqrt{\dfrac{144}{25}} = \dfrac{12}{5}\)
- सरल गर्नुहोस् (Simplify): \((1 - \sin A)(1 + \sin A) \sec^2 A\)
The simplification is
\((1 - \sin A)(1 + \sin A) \sec^2 A\)
or\((1 - \sin^2 A )\sec^2 A\)
or\(\cos^2 A \cdot \sec^2 A\)
or\( 1\)
- गुणनखण्ड गर्नुहोस् (Factorize): \(\cos^4 A - 2\cos^2 A - 8\)
Given that
\(\cos^4 A - 2\cos^2 A - 8\)
or\(\cos^4 A - (4\cos^2 A -2\cos^2 A)- 8\)
or\(\cos^4 A - 4\cos^2 A +2\cos^2 A- 8\)
or\(\cos^2 A(\cos^2 A - 4) +2(\cos^2 A- 4)\)
or\((\cos^2 A - 4) (\cos^2 A+2)\)
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{\sec\theta \cdot \cos\theta}{\csc\theta} = \sin\theta\)
Given that
\(\dfrac{\sec\theta \cdot \cos\theta}{\csc\theta}\)
or\(\dfrac{1}{\csc\theta}\)
or\(\sin\theta\)
- प्रमाणित गर्नुहोस् (Prove that): \((\sin X + \cos X)^2 = 1 + 2\sin X \cos X\)
Given that
\((\sin X + \cos X)^2\)
or\(\sin^2 X + 2\sin X \cos X + \cos^2 X\)
or\((\sin^2 X + \cos^2 X) + 2\sin X \cos X\)
or\(1 + 2\sin X \cos X\)
- प्रमाणित गर्नुहोस् (Prove that): \(\sqrt{1 - \cos^2\theta} \cdot \sec\theta = \tan\theta\)
Given that
\(\sqrt{1 - \cos^2\theta} \cdot \sec\theta\)
or\(\sqrt{\sin^2\theta} \cdot \sec\theta\)
or\(\sin\theta \cdot \dfrac{1}{\cos\theta}\)
or\(\tan\theta\)
- प्रमाणित गर्नुहोस् (Prove that): \(\sin^2\beta + \sin^2\beta \cdot \cot^2\beta = 1\)
Given that
\(\sin^2\beta + \sin^2\beta \cdot \cot^2\beta\)
or\(\sin^2\beta (1 + \cot^2\beta)\)
or\(\sin^2\beta \cdot \csc^2\beta\)
or\(\sin^2\beta \cdot \dfrac{1}{\sin^2\beta}\)
or\(1\)
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{1}{\tan\theta + \cot\theta} = \sin\theta \cdot \cos\theta\)
Given that
\(\dfrac{1}{\tan\theta + \cot\theta}\)
or\(\dfrac{1}{\dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta}}\)
or\(\dfrac{1}{\dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta}}\)
or\(\dfrac{1}{\dfrac{1}{\sin\theta \cos\theta}}\)
or\(\sin\theta \cdot \cos\theta\)
- प्रमाणित गर्नुहोस् (Prove that): \(\cos^2\alpha + \sin^2\alpha \cdot \cot^2\alpha = 2\cos^2\alpha\)
Given that
\(\cos^2\alpha + \sin^2\alpha \cdot \cot^2\alpha\)
or\(\cos^2\alpha + \sin^2\alpha \cdot \dfrac{\cos^2\alpha}{\sin^2\alpha}\)
or\(\cos^2\alpha + \cos^2\alpha\)
or\(2\cos^2\alpha\)
Group 'C' in BLE (4 Marks Each)
- यदि \(\csc \theta = \dfrac{b}{a}\) भए, सिद्ध गर्नुहोस्।
(If \(\csc \theta = \dfrac{b}{a}\), prove that): \(\sqrt{b^2 - a^2} = a \cot \theta\)
Given that
\(\csc \theta = \dfrac{b}{a}=\dfrac{\text{hypotenuse}}{\text{opposite}}\)
In a right triangle
\(base = \sqrt{h^2 - p^2}= \sqrt{b^2 - a^2}\)
Now
\(\cot \theta = \dfrac{\text{adjacent}}{\text{opposite}}\)
or\(\cot \theta = \dfrac{\sqrt{b^2 - a^2}}{a}\)
or\(\sqrt{b^2 - a^2} = a \cot \theta\)
- समकोणी त्रिभुज ABC मा \(\angle B = 90^\circ\), \(\sin A = \dfrac{3}{5}\) भए, \(\cos A\) र \(\tan A\) पत्ता लगाउनुहोस्।
(In a right-angled triangle ABC, if \(\angle B = 90^\circ\) and \(\sin A = \dfrac{3}{5}\), find \(\cos A\) and \(\tan A\).)
Given that in \(\triangle ABC\), \(\angle B = 90^\circ\) and \(\sin A = \dfrac{3}{5}\).
or\(\sin A = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{3}{5}\)
orSo, \(BC = 3\), \(AC = 5\)
By Pythagoras theorem, we have
\(AB = \sqrt{AC^2 - BC^2} = \sqrt{25 - 9} = \sqrt{16} = 4\)
Thus
\(\cos A = \dfrac{AB}{AC} = \dfrac{4}{5}\)
\(\tan A = \dfrac{BC}{AB} = \dfrac{3}{4}\)
- दिएको चित्रअनुसार \(\sin\theta\), \(\cos\theta\) र \(\tan\theta\) को मान पत्ता लगाउनुहोस्।
(Find the values of \(\sin\theta\), \(\cos\theta\) and \(\tan\theta\) from the given figure.)
In right-angled triangle \(ABC\), right angle is at \(B\), and \(\angle A = \theta\).
Given: \(BC = 5\), \(AC = 13\).
By Pythagoras theorem, we have
\(AB = \sqrt{AC^2 - BC^2} = \sqrt{169 - 25} = \sqrt{144} = 12\)
Thus
\(\sin\theta = \dfrac{BC}{AC} = \dfrac{5}{13}\)
\(\cos\theta = \dfrac{AB}{AC} = \dfrac{12}{13}\)
\(\tan\theta = \dfrac{BC}{AB} = \dfrac{5}{12}\)
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{\sin^2 A}{\sin A - \cos A} + \dfrac{\cos^2 A}{\cos A - \sin A} = \sin A + \cos A\)
Given that
\(\dfrac{\sin^2 A}{\sin A - \cos A} + \dfrac{\cos^2 A}{\cos A - \sin A}\)
or\(\dfrac{\sin^2 A}{\sin A - \cos A} - \dfrac{\cos^2 A}{\sin A - \cos A}\)
or\(\dfrac{\sin^2 A - \cos^2 A}{\sin A - \cos A}\)
or\(\dfrac{(\sin A - \cos A)(\sin A + \cos A)}{\sin A - \cos A}\)
or\(\sin A + \cos A\)
- प्रमाणित गर्नुहोस् (Prove that): \((\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2 = 2\)
Given that
\((\sin \theta + \cos \theta)^2 + (\sin \theta - \cos \theta)^2\)
or\(\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta\)
or\((\sin^2 \theta + \cos^2 \theta) + (\sin^2 \theta + \cos^2 \theta)\)
or\(1 + 1\)
or\(2\)
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta\)
Given that
\(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta}\)
or\(\dfrac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}\)
or\(\dfrac{\sin^2 \theta + 1 + 2\cos \theta + \cos^2 \theta}{\sin \theta (1 + \cos \theta)}\)
or\(\dfrac{(\sin^2 \theta + \cos^2 \theta) + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
or\(\dfrac{1 + 1 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}\)
or\(\dfrac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)}\)
or\(\dfrac{2}{\sin \theta} \)
or\(2 \csc \theta\)
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{\cos A}{1 + \sin A} + \dfrac{\cos A}{1 - \sin A} = 2 \sec A\)
Given that
\(\dfrac{\cos A}{1 + \sin A} + \dfrac{\cos A}{1 - \sin A}\)
or\(\cos A \left( \dfrac{1}{1 + \sin A} + \dfrac{1}{1 - \sin A} \right)\)
or\(\cos A \left( \dfrac{(1 - \sin A) + (1 + \sin A)}{(1 + \sin A)(1 - \sin A)} \right)\)
or\(\cos A \left( \dfrac{2}{1 - \sin^2 A} \right)\)
or\(\cos A \cdot \dfrac{2}{\cos^2 A}\)
or\(\dfrac{2}{\cos A}\)
or\(2 \sec A\)
- प्रमाणित गर्नुहोस् (Prove that): \((a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2 = a^2 + b^2\)
Given that
\((a \cos \theta + b \sin \theta)^2 + (a \sin \theta - b \cos \theta)^2\)
or\(a^2 \cos^2 \theta + 2ab \sin \theta \cos \theta + b^2 \sin^2 \theta + a^2 \sin^2 \theta - 2ab \sin \theta \cos \theta + b^2 \cos^2 \theta\)
or\(a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta)\)
or\(a^2(1) + b^2(1)\)
or\(a^2 + b^2\)
- प्रमाणित गर्नुहोस् (Prove that): \(\sqrt{\dfrac{1 + \cos A}{1 - \cos A}} = \csc A + \cot A\)
Given that
\(\sqrt{\dfrac{1 + \cos A}{1 - \cos A}}\)
or\(\sqrt{\dfrac{(1 + \cos A)}{(1 - \cos A)} \times \dfrac{(1 + \cos A)}{(1 + \cos A)}}\)
or\( \sqrt{\dfrac{(1 + \cos A)^2}{1 - \cos^2 A}}\)
or\(\sqrt{\dfrac{(1 + \cos A)^2}{\sin^2 A}} \)
or\(\dfrac{1 + \cos A}{\sin A}\)
or\(\dfrac{1}{\sin A} + \dfrac{\cos A}{\sin A}\)
or\(\csc A + \cot A\)
- प्रमाणित गर्नुहोस् (Prove that): \(\sqrt{\dfrac{1 - \sin \theta}{1 + \sin \theta}} = \sec \theta - \tan \theta\)
Given that
\(\sqrt{\dfrac{1 - \sin \theta}{1 + \sin \theta}}\)
or\(\sqrt{\dfrac{(1 - \sin \theta)}{(1 + \sin \theta)} \times \dfrac{(1 - \sin \theta)}{(1 - \sin \theta)}}\)
or\(\sqrt{\dfrac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}}\)
or\(\sqrt{\dfrac{(1 - \sin \theta)^2}{\cos^2 \theta}}\)
or\(\dfrac{1 - \sin \theta}{\cos \theta}\)
or\(\dfrac{1}{\cos \theta} - \dfrac{\sin \theta}{\cos \theta}\)
or\(\sec \theta - \tan \theta\)
- प्रमाणित गर्नुहोस् (Prove that): \(\csc^4 \theta - \csc^2 \theta = \cot^2 \theta + \cot^4 \theta\)
Given that
\(\csc^4 \theta - \csc^2 \theta\)
or\(\csc^2 \theta (\csc^2 \theta - 1)\)
or\(\csc^2 \theta \cdot \cot^2 \theta\)
or\((1 + \cot^2 \theta) \cdot \cot^2 \theta\)
or\(\cot^2 \theta + \cot^4 \theta\)
- यदि \(\sin A - \cos A = 0\) भए, \(\sin A\), \(\cos A\) र \(\tan A\) को मान निकाल्नुहोस्।
(If \(\sin A - \cos A = 0\), find the values of \(\sin A\), \(\cos A\) and \(\tan A\).)
Given that
\(\sin A - \cos A = 0\)
or\(\sin A = \cos A\)
or\(\dfrac{\sin A}{\cos A} = 1\)
or\(\tan A = 1\)
By Pythagoras theorem, we have
\(h = \sqrt{p^2 + b^2} = \sqrt{1 +1} = \sqrt{2} \)
Thus
\(\sin A = \dfrac{p}{h} = \dfrac{1}{\sqrt{2}}\)
\(\cos A = \dfrac{b}{h} = \dfrac{1}{\sqrt{2}}\)
\(\tan A= 1\)
- यदि \(b \cot \theta = a\) भए, \(\dfrac{a \cos \theta - b \sin \theta}{a \cos \theta + b \sin \theta}\) को मान निकाल्नुहोस्।
(If \(b \cot \theta = a\), find the value of \(\dfrac{a \cos \theta - b \sin \theta}{a \cos \theta + b \sin \theta}\).
Given that
\(b \cot \theta = a\)
or\(\cot \theta = \dfrac{a}{b}=\dfrac{\cos \theta}{\sin \theta}\)
Thus
\(\dfrac{a \cos \theta - b \sin \theta}{a \cos \theta + b \sin \theta}\)
or\(\dfrac{a a - b b}{a a + b b}\)
or\(\dfrac{a^2 - b^2}{a^2 + b^2}\) - यदि \(3 \cot \theta = 4\) भए, \(\dfrac{3 \cos \theta + 2 \sin \theta}{3 \cos \theta - 2 \sin \theta}\) को मान निकाल्नुहोस्।
(If \(3 \cot \theta = 4\), find the value of \(\dfrac{3 \cos \theta + 2 \sin \theta}{3 \cos \theta - 2 \sin \theta}\).)
Given that
\(3 \cot \theta = 4\)
or\(\cot \theta = \dfrac{4}{3} = \dfrac{\cos \theta}{\sin \theta}\)
Thus
\(\dfrac{3 \cos \theta + 2 \sin \theta}{3 \cos \theta - 2 \sin \theta}\)
or\(\dfrac{3 \cdot 4 + 2 \cdot 3}{3 \cdot 4 - 2 \cdot 3}\)
or\(\dfrac{12 + 6}{12 - 6}\)
or\(\dfrac{18}{6} = 3\)
Standard Angles and Complementary Angles
Group 'B' in BLE (2 Marks Each)
- मान निकाल्नुहोस्। (Find the value of): \(\sin^2 90^\circ + \cos^2 60^\circ + \cos^2 45^\circ + \sin 30^\circ\)
Given that
\(\sin^2 90^\circ + \cos^2 60^\circ + \cos^2 45^\circ + \sin 30^\circ\)
or\((1)^2 + \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 + \dfrac{1}{2}\)
or\(1 + \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{1}{2}\)
or\(1 + \dfrac{1}{4} + 1\)
or\(\dfrac{9}{4}\)
- मान निकाल्नुहोस्। (Find the value of): \(\sin^2 45^\circ + \cos^2 30^\circ - \tan^2 45^\circ\)
Given that
\(\sin^2 45^\circ + \cos^2 30^\circ - \tan^2 45^\circ\)
or\(\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 - (1)^2\)
or\(\dfrac{1}{2} + \dfrac{3}{4} - 1\)
or\(\dfrac{2}{4} + \dfrac{3}{4} - \dfrac{4}{4}\)
or\(\dfrac{1}{4}\)
- मान निकाल्नुहोस्। (Find the value of): \(\sin \dfrac{\pi^c}{6} + \cos \dfrac{\pi^c}{3} + \tan \dfrac{\pi^c}{4}\)
Given that
\(\sin \dfrac{\pi}{6} + \cos \dfrac{\pi}{3} + \tan \dfrac{\pi}{4}\)
or\(\dfrac{1}{2} + \dfrac{1}{2} + 1\)
or\(1 + 1\)
or\(2\)
- यदि \(A = 60^\circ\), \(B = 30^\circ\) भए मान निकाल्नुहोस्।
(If \(A = 60^\circ\), \(B = 30^\circ\), find the value of): \(\cos A \sin B + \sin A \cos B\)
Given that
\(\cos A \sin B + \sin A \cos B\)
or\(\cos 60^\circ \sin 30^\circ + \sin 60^\circ \cos 30^\circ\)
or\(\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right) + \left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)\)
or\(\dfrac{1}{4} + \dfrac{3}{4}\)
or\(1\)
- यदि \(A = 60^\circ\), \(B = 45^\circ\) र \(C = 30^\circ\) भए प्रमाणित गर्नुहोस्: \(4 \cos A + 2\sqrt{2} \cos B + 2\sqrt{3} \cos C = 7\)
Given that
\(4 \cos A + 2\sqrt{2} \cos B + 2\sqrt{3} \cos C\)
or\(4 \cos 60^\circ + 2\sqrt{2} \cos 45^\circ + 2\sqrt{3} \cos 30^\circ\)
or\(4 \cdot \dfrac{1}{2} + 2\sqrt{2} \cdot \dfrac{1}{\sqrt{2}} + 2\sqrt{3} \cdot \dfrac{\sqrt{3}}{2}\)
or\(2 + 2 + 3\)
or\(7\)
- यदि \(A = 60^\circ\), \(B = 45^\circ\) र \(C = 90^\circ\) भए, \(\tan^2 A + 2 \sin^2 B + \sin^2 C\) को मान निकाल्नुहोस्।
Given that
\(\tan^2 A + 2 \sin^2 B + \sin^2 C\)
or\(\tan^2 60^\circ + 2 \sin^2 45^\circ + \sin^2 90^\circ\)
or\((\sqrt{3})^2 + 2 \left(\dfrac{1}{\sqrt{2}}\right)^2 + (1)^2\)
or\(3 + 2 \cdot \dfrac{1}{2} + 1\)
or\(3 + 1 + 1\)
or\(5\)
- यदि \( A = 60^\circ \) र \( B = 30^\circ \) भए, प्रमाणित गर्नुहोस्: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
Given that
\(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
or\(\cos(60^\circ + 30^\circ) = \cos 60^\circ \cos 30^\circ - \sin 60^\circ \sin 30^\circ\)
or\(\cos 90^\circ = \left(\frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)\)
or\(0 = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}\)
or\(0 = 0\)
Hence, the identity is proved. - यदि \( A = 60^\circ \) भए, प्रमाणित गर्नुहोस्: \(\sin 2A = 2 \sin A \cos A\)
Given that
\(\sin 2A = 2 \sin A \cos A\)
or\(\sin(2 \cdot 60^\circ) = 2 \sin 60^\circ \cos 60^\circ\)
or\(\sin 120^\circ = 2 \left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)\)
or\(\frac{\sqrt{3}}{2} = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}\)
or\(\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}\)
Hence, the identity is proved. - यदि \( A = 30^\circ \) भए, प्रमाणित गर्नुहोस्: \(\cos 2A = 2 \cos^2 A - 1\)
Given that
\(\cos 2A = 2 \cos^2 A - 1\)
or\(\cos(2 \cdot 30^\circ) = 2 \cos^2 30^\circ - 1\)
or\(\cos 60^\circ = 2 \left(\frac{\sqrt{3}}{2}\right)^2 - 1\)
or\(\frac{1}{2} = 2 \cdot \frac{3}{4} - 1\)
or\(\frac{1}{2} = \frac{6}{4} - 1\)
or\(\frac{1}{2} = \frac{3}{2} - 1\)
or\(\frac{1}{2} = \frac{1}{2}\)
Hence, the identity is proved. - \(x\) को मान निकाल्नुहोस्: \(\sin 3x = \cos 2x\)
Given that
\(\sin 3x = \cos 2x\)
or\(3x+2x=90\)
or\(5x=90\)
or\(x=18^0\)
- \(x\) को मान निकाल्नुहोस्: \(\sin 6x = \cos 3x\)
Given that
\(\sin 6x = \cos 3x\)
or\(6x+3x=90\)
or\(9x=90\)
or\(x=10^0\)
- प्रमाणित गर्नुहोस्: \(\frac{\sin A \cdot \sin(90^\circ - A) \cdot \tan(90^\circ - A)}{\cos(90^\circ - A) \cdot \cot A} = \cos A\)
Given that
\(\frac{\sin A \cdot \sin(90^\circ - A) \cdot \tan(90^\circ - A)}{\cos(90^\circ - A) \cdot \cot A} = \cos A\)
or\(\frac{\sin A \cdot \cos A \cdot \cot A}{\sin A \cdot \cot A} = \cos A\)
or\(\cos A = \cos A\)
Hence, the identity is proved. - प्रमाणित गर्नुहोस्: \(\cos(90^\circ - A) \cdot \text{cosec}(90^\circ - A) = \tan A\)
Given that
\(\cos(90^\circ - A) \cdot \text{cosec}(90^\circ - A) = \tan A\)
or\(\sin A \cdot \sec A = \tan A\)
or\(\frac{\sin A}{\cos A} = \tan A\)
or\(\tan A = \tan A\)
Hence, the identity is proved. - प्रमाणित गर्नुहोस्: \(\cos 25^\circ + \sin 55^\circ = \sin 65^\circ + \cos 35^\circ\)
LHS
\(\cos 25^\circ + \sin 55^\circ \)
or\( \sin (90^\circ - 65^\circ) + \cos (90^\circ - 35^\circ)\)
or\(\sin 65^\circ + \cos 35^\circ \)
Hence, the identity is proved. - यदि \( A = 45^\circ \) भए, प्रमाणित गर्नुहोस्: \(\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}\)
Given that
\(\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}\)
or\(\tan (2 \cdot 45^\circ) = \frac{2 \tan 45^\circ}{1 - \tan^2 45^\circ}\)
or\(\tan 90^\circ = \frac{2 \cdot 1}{1 - 1^2}\)
or\(\infty = \frac{2}{0}\)
or\(\infty = \infty \)
Group 'C' in BLE (4 Marks Each)
- \(30^\circ\) का सबै त्रिकोणमितीय अनुपातहरू ज्यामितीय विधिबाट पत्ता लगाउनुहोस्। (Find all the trigonometric ratios of \(30^\circ\) geometrically.)
Consider an equilateral triangle \(ABC\) with each side of length \(2a\) units. All interior angles are \(60^\circ\).
Drop a perpendicular from vertex \(A\) to side \(BC\), and let the foot of the perpendicular be point \(D\).
In tjis case
\(BD = DC = a\) Also, AD bisects the angle at \(A\), so \(\angle BAD = 30^\circ\). Thus, triangle \(ABD\) is a right triangle with
right angle at \(D\)
hypotenuse \(AB = 2a\)
side opposite \(30^\circ\): \(BD = a\)
side adjacent to \(30^\circ\): \(AD = ?\)
Using the Pythagorean Theorem, in right triangle \(ABD\), we get \(AD=\sqrt{AB^2-BD^2}= \sqrt{(2a)^2-a^2}=\sqrt{3a^2}=a\sqrt{3}\)
Now, using the definitions of trigonometric functions in a right triangle, we get \(\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}\)
\(\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \)
\(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} \)
\(\csc(30^\circ) = \frac{1}{\sin(30^\circ)} = 2\)
\(\sec(30^\circ) = \frac{1}{\cos(30^\circ)} = \frac{2}{\sqrt{3}} \)
\(\cot(30^\circ) = \frac{1}{\tan(30^\circ)} = \sqrt{3}\) - \(45^\circ\) का सबै त्रिकोणमितीय अनुपातहरू ज्यामितीय विधिबाट पत्ता लगाउनुहोस्। (Find all the trigonometric ratios of \(45^\circ\) geometrically.)
Consider an isosceles right-angled triangle \(PQR\) with \(\angle Q = 90^\circ\) and the two legs equal: \(PQ = QR = a\) units.
Since the triangle is isosceles and right-angled, the other two angles are equal:
\(\angle P = \angle R = 45^\circ\)
Using the Pythagorean Theorem, the hypotenuse is:
\(PR = \sqrt{PQ^2 + QR^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}\) Now, consider the reference angle \(45^\circ\) at vertex \(P\). In right triangle \(PQR\):
right angle at \(Q\)
hypotenuse \(PR = a\sqrt{2}\)
side opposite \(45^\circ\) (at \(P\)): \(QR = a\)
side adjacent to \(45^\circ\): \(PQ = a\)
Using the definitions of trigonometric ratios: \(\sin(45^\circ) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a}{a\sqrt{2}} = \dfrac{1}{\sqrt{2}} \)
\(\cos(45^\circ) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{a}{a\sqrt{2}} = \dfrac{1}{\sqrt{2}} \)
\(\tan(45^\circ) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a}{a} = 1\)
\(\csc(45^\circ) = \dfrac{1}{\sin(45^\circ)} = \sqrt{2}\)
\(\sec(45^\circ) = \dfrac{1}{\cos(45^\circ)} = \sqrt{2}\)
\(\cot(45^\circ) = \dfrac{1}{\tan(45^\circ)} = 1\) - \(60^\circ\) का सबै त्रिकोणमितीय अनुपातहरू ज्यामितीय विधिबाट पत्ता लगाउनुहोस्। (Find all the trigonometric ratios of \(60^\circ\) geometrically.)
Consider an equilateral triangle \(ABC\) with each side of length \(2a\) units. All interior angles are \(60^\circ\).
Drop a perpendicular from vertex \(B\) to side \(AC\), and let the foot of the perpendicular be point \(E\).
In this case
\(AE = EC = a\) Also, \(BE\) is an altitude and median, so it bisects side \(AC\) and angle \(B\). Thus, in right triangle \(BEC\):
right angle at \(E\)
hypotenuse \(BC = 2a\)
side adjacent to \(60^\circ\) (at \(C\)): \(EC = a\)
side opposite \(60^\circ\): \(BE = a\sqrt{3}\)
Using the Pythagorean Theorem to confirm: \(BE = \sqrt{BC^2 - EC^2} = \sqrt{(2a)^2 - a^2} = \sqrt{3a^2} = a\sqrt{3}\)
Now, using the definitions of trigonometric ratios in right triangle \(BEC\) with reference angle \(60^\circ\) at \(C\)
\(\sin(60^\circ) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a\sqrt{3}}{2a} = \dfrac{\sqrt{3}}{2}\)
\(\cos(60^\circ) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{a}{2a} = \dfrac{1}{2}\)
\(\tan(60^\circ) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a\sqrt{3}}{a} = \sqrt{3}\)
\(\csc(60^\circ) = \dfrac{1}{\sin(60^\circ)} = \dfrac{2}{\sqrt{3}} \)
\(\sec(60^\circ) = \dfrac{1}{\cos(60^\circ)} = \dfrac{1}{\tfrac{1}{2}} = 2\)
\(\cot(60^\circ) = \dfrac{1}{\tan(60^\circ)} = \dfrac{1}{\sqrt{3}} \) - \(0^\circ\) का सबै त्रिकोणमितीय अनुपातहरू ज्यामितीय विधिबाट पत्ता लगाउनुहोस्। (Find all the trigonometric ratios of \(0^\circ\) geometrically.)
Consider a right-angled triangle \(ABC\) with right angle at \(B\), and let \(C\) moves toward B, such that angle at \(A\) becoms smaller and smaller, approaching \(0^\circ\).
In the limiting case when \(\angle A = 0^\circ\):
– Point \(C\) moves down and coincides with point \(B\).
– Side opposite to \(\angle A\) (i.e., \(BC\)) becomes \(0\).
– Side adjacent to \(\angle A\) (i.e., \(AB\)) becomes equal to the hypotenuse \(AC\).
Assume the hypotenuse \(AC = a\). Then in the limiting position
Opposite side = \(BC = 0\)
Adjacent side = \(AB = a\)
Hypotenuse = \(AC = a\) Using the definitions of trigonometric ratios in this limiting right triangle with reference angle \(0^\circ\) at \(A\)
\(\sin(0^\circ) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{0}{a} = 0\)
\(\cos(0^\circ) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{a}{a} = 1\)
\(\tan(0^\circ) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{0}{a} = 0\)
\(\csc(0^\circ) = \dfrac{1}{\sin(0^\circ)} = \dfrac{1}{0} \to \text{undefined}\)
\(\sec(0^\circ) = \dfrac{1}{\cos(0^\circ)} = \dfrac{1}{1} = 1\)
\(\cot(0^\circ) = \dfrac{1}{\tan(0^\circ)} = \dfrac{1}{0} \to \text{undefined}\) - \(90^\circ\) का सबै त्रिकोणमितीय अनुपातहरू ज्यामितीय विधिबाट पत्ता लगाउनुहोस्। (Find all the trigonometric ratios of \(90^\circ\) geometrically.)
Consider a right-angled triangle \(ABC\) with right angle at \(B\), and let \(A\) moves toward B, such that angle at \(A\) becoms larger and larger, approaching \(90^\circ\).
In the limiting case when \(\angle A = 90^\circ\):
– Point \(A\) moves down and coincides with point \(B\).
– Side opposite to \(\angle A\) (i.e., \(BC\)) becomes equal to the hypotenuse \(AC\).
– Side adjacent to \(\angle A\) (i.e., \(AB =0\))
Assume the hypotenuse \(AC = a\). Then in the limiting position
Opposite side = \(BC = a\)
Adjacent side = \(AB = 0\)
Hypotenuse = \(AC = a\) Using the definitions of trigonometric ratios in this limiting right triangle with reference angle \(0^\circ\) at \(A\)
\(\sin(90^\circ) = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a}{a} =1\)
\(\cos(90^\circ) = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{0}{a} = 0\)
\(\tan(90^\circ) = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a}{0} = \infty\)
\(\csc(90^\circ) =\dfrac{\text{hypotenuse}}{\text{opposite}}\dfrac{a}{a} =1\)
\(\sec(90^\circ) = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{a}{0} = \infty\)
\(\cot(90^\circ) =\dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{0}{a} = 0\) - यदि \(A = 30^\circ\) र \(B = 60^\circ\) भए, प्रमाणित गर्नुहोस्: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
The LHS
\(\cos(A + B) = \cos(30^\circ + 60^\circ) = \cos 90^\circ = 0 \)
The RHS
We know that
\(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
\(\cos 60^\circ = \frac{1}{2}\)
\(\sin 30^\circ = \frac{1}{2}\)
\(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
Hence
\(\cos A \cos B - \sin A \sin B \)
\(\cos 30^\circ \cos 60^\circ - \sin 30^\circ \sin 60^\circ \)
\(\left( \frac{\sqrt{3}}{2} \right) \left( \frac{1}{2} \right) - \left( \frac{1}{2} \right) \left( \frac{\sqrt{3}}{2} \right)\)
\(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \)
\(0\)
Here
\(LHS = RHS \)
- यदि \(A = 30^\circ\) र \(B = 60^\circ\) भए, प्रमाणित गर्नुहोस्: \(\sin(A + B) = \sin A \cos B + \cos A \sin B\)
The LHS
\(\sin(A + B) = \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1\)
The RHS
We know that
\(\sin 30^\circ = \dfrac{1}{2}\)
\(\cos 60^\circ = \dfrac{1}{2}\)
\(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\)
\(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\)
Hence
\(\sin A \cos B + \cos A \sin B\)
\(\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ\)
\(\left( \dfrac{1}{2} \right) \left( \dfrac{1}{2} \right) + \left( \dfrac{\sqrt{3}}{2} \right) \left( \dfrac{\sqrt{3}}{2} \right)\)
\(\dfrac{1}{4} + \dfrac{3}{4}\)
\(1\)
Here
\(LHS = RHS\)
- यदि \(A = 60^\circ\), \(B = 45^\circ\) र \(C = 30^\circ\) भए, प्रमाणित गर्नुहोस्: \(\tan^2 A + 4 \cos^2 B + 3 \sec^2 C = 9\)
We are given:
\(A = 60^\circ\), \(B = 45^\circ\), \(C = 30^\circ\)
First, recall the following standard values:
\(\tan 60^\circ = \sqrt{3}\)
\(\cos 45^\circ = \dfrac{1}{\sqrt{2}}\)
\(\sec 30^\circ = \dfrac{2}{\sqrt{3}}\)
Now compute each term:
1. \(\tan^2 A = \tan^2 60^\circ = (\sqrt{3})^2 = 3\)
2. \(4 \cos^2 B = 4 \cos^2 45^\circ = 4 \left( \dfrac{1}{\sqrt{2}} \right)^2 = 4 \left( \dfrac{1}{2} \right) = 2\)
3. \(3 \sec^2 C = 3 \sec^2 30^\circ = 3 \left( \dfrac{2}{\sqrt{3}} \right)^2 = 3 \left( \dfrac{4}{3} \right) = 4\)
Now add them together:
\(\tan^2 A + 4 \cos^2 B + 3 \sec^2 C = 3 + 2 + 4 = 9\)
Hence, the given identity is verified.
LHS = RHS = 9
- यदि \(\theta = 30^\circ\) र \(\alpha = 60^\circ\) भए, प्रमाणित गर्नुहोस्: \(\tan(\alpha - \theta) = \dfrac{\tan \alpha - \tan \theta}{1 + \tan \alpha \tan \theta}\)
The LHS
\(\tan(\alpha - \theta) = \tan(60^\circ - 30^\circ) = \tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
The RHS
We know that
\(\tan 60^\circ = \sqrt{3}\)
\(\tan 30^\circ = \dfrac{1}{\sqrt{3}}\)
Hence,
\(\dfrac{\tan \alpha - \tan \theta}{1 + \tan \alpha \tan \theta} = \dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\)
\(= \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \left( \sqrt{3} \cdot \dfrac{1}{\sqrt{3}} \right)}\)
\(= \dfrac{\dfrac{3 - 1}{\sqrt{3}}}{1 + 1}\)
\(= \dfrac{\dfrac{2}{\sqrt{3}}}{2} = \dfrac{2}{\sqrt{3}} \cdot \dfrac{1}{2} = \dfrac{1}{\sqrt{3}}\)
Here,
LHS = RHS = \(\dfrac{1}{\sqrt{3}}\)
- प्रमाणित गर्नुहोस् (Prove that): \(\sqrt{33 + \sqrt{7 + \sqrt{8 \sin 30^\circ}}} = 6\)
We begin from the innermost expression and simplify step by step.
We know:
\(\sin 30^\circ = \dfrac{1}{2}\)
Step 1: Evaluate \(\sqrt{8 \sin 30^\circ}\)
\(\sqrt{8 \cdot \dfrac{1}{2}} = \sqrt{4} = 2\)
Step 2: Evaluate \(\sqrt{7 + \sqrt{8 \sin 30^\circ}} = \sqrt{7 + 2} = \sqrt{9} = 3\)
Step 3: Evaluate the entire expression
\(\sqrt{33 + \sqrt{7 + \sqrt{8 \sin 30^\circ}}} = \sqrt{33 + 3} = \sqrt{36} = 6\)
Hence, the given equation is verified.
LHS = RHS = 6
- यदि \(A = 30^\circ\) भए, प्रमाणित गर्नुहोस्: (If \(A = 30^\circ\), prove that): \(\sin 3A = 3 \sin A - 4 \sin^3 A\)
The LHS
\(\sin 3A = \sin(3 \times 30^\circ) = \sin 90^\circ = 1\)
The RHS
\(\sin A = \sin 30^\circ = \dfrac{1}{2}\)
Hence
\(3 \sin A - 4 \sin^3 A = 3 \left( \dfrac{1}{2} \right) - 4 \left( \dfrac{1}{2} \right)^3\)
\(= \dfrac{3}{2} - 4 \left( \dfrac{1}{8} \right)\)
\(= \dfrac{3}{2} - \dfrac{4}{8} = \dfrac{3}{2} - \dfrac{1}{2} = 1\)
Here,
LHS = RHS = 1
- प्रमाणित गर्नुहोस् (Prove that): \(\sec \theta \cdot \csc(90^\circ - \theta) - \tan \theta \cdot \cot(90^\circ - \theta) = 1\)
We use the following co-function identities:
\(\csc(90^\circ - \theta) = \sec \theta\)
\(\cot(90^\circ - \theta) = \tan \theta\)
Now substitute into the left-hand side (LHS):
\(\sec \theta \cdot \csc(90^\circ - \theta) - \tan \theta \cdot \cot(90^\circ - \theta)\)
or\( \sec \theta \cdot \sec \theta - \tan \theta \cdot \tan \theta\)
or\( \sec^2 \theta - \tan^2 \theta\)
or\( 1\)
Therefore,
LHS = 1 = RHS
- प्रमाणित गर्नुहोस् (Prove that): \(\csc(90^\circ - \theta) \cdot \sec \theta - \cot(90^\circ - \theta) \cdot \tan \theta = 1\)
We use the following co-function identities:
\(\csc(90^\circ - \theta) = \sec \theta\)
\(\cot(90^\circ - \theta) = \tan \theta\)
Now substitute into the left-hand side (LHS):
\(\csc(90^\circ - \theta) \cdot \sec \theta - \cot(90^\circ - \theta) \cdot \tan \theta\)
or\(\sec \theta \cdot \sec \theta - \tan \theta \cdot \tan \theta\)
or\(\sec^2 \theta - \tan^2 \theta\)
or\(1\)
Therefore,
LHS = 1 = RHS
- प्रमाणित गर्नुहोस् (Prove that): \(\dfrac{\sin(90^\circ - \theta) \cdot \sec(90^\circ - \theta) \cdot \csc(90^\circ - \theta)}{\cot(90^\circ - \theta) \cdot \tan(90^\circ - \theta) \cdot \cos(90^\circ - \theta)} = \csc^2\theta\)
We use the following co-function identities:
\(\sin(90^\circ - \theta) = \cos \theta\)
\(\cos(90^\circ - \theta) = \sin \theta\)
\(\sec(90^\circ - \theta) = \csc \theta\)
\(\csc(90^\circ - \theta) = \sec \theta\)
\(\tan(90^\circ - \theta) = \cot \theta\)
\(\cot(90^\circ - \theta) = \tan \theta\)
Now substitute all into the LHS:
Numerator:
\(\sin(90^\circ - \theta) \cdot \sec(90^\circ - \theta) \cdot \csc(90^\circ - \theta)\)
or\( \cos \theta \cdot \csc \theta \cdot \sec \theta\)
or\( \csc \theta \)(i)
Denominator:
\(\cot(90^\circ - \theta) \cdot \tan(90^\circ - \theta) \cdot \cos(90^\circ - \theta)\)
or\( \tan \theta \cdot \cot \theta \cdot \sin \theta\)
or\( \sin \theta\)(ii)
Thus, the entire expression is
\(\dfrac{\csc \theta}{\sin \theta} = \csc^2 \theta\)
Therefore,
LHS = RHS
Solution of Right Angled Triangle, Height and Distance
Group 'C' in BLE (4 Marks Each)
- \(\triangle PQR\) को हल गर्नुहोस् जसमा \(\angle R = 90^\circ\), \(PQ = 4\sqrt{3} \text{ cm}\) र \(QR = 2\sqrt{3} \text{ cm}\) छन्। (Solve the \(\triangle PQR\) where \(\angle R = 90^\circ\), \(PQ = 4\sqrt{3} \text{ cm}\) and \(QR = 2\sqrt{3} \text{ cm}\).)
In the given right angled triangle
\(PR=\sqrt{h^2-p^2}=\sqrt{(4 \sqrt{3})^2-(2 \sqrt{3})^2 }=\sqrt{48-12}=\sqrt{36}\)
or\(PR=6cm\)
Also
\(\sin \theta=\frac{p}{h}=\frac{2 \sqrt{3}}{4 \sqrt{3}}=\frac{1}{2}\)
or\(\theta=60^0\)
Hence, the solution of the triangle is
\(\measuredangle P=60^0, \measuredangle Q=30^0, PR=6cm\)
- एउटा समकोणी त्रिभुज ABC मा \(\angle A = 30^\circ\), \(\angle B = 90^\circ\) र \(b = 4\sqrt{3} \text{ cm}\) भए, \(\triangle ABC\) को हल गर्नुहोस्। (In a right-angled triangle ABC, \(\angle A = 30^\circ\), \(\angle B = 90^\circ\) and \(b = 4\sqrt{3} \text{ cm}\), solve the \(\triangle ABC\).)
In the given right angled triangle
\(\measuredangle C=60^0\)
Next
\(\sin 30=\frac{BC}{4 \sqrt{3}}\)
or\(\frac{1}{2}=\frac{BC}{4 \sqrt{3}}\)
or\(BC=2 \sqrt{3}\) cm
Again
\(\cos 30=\frac{AB}{4 \sqrt{3}}\)
or\(\frac{\sqrt{3}}{2}=\frac{BC}{4 \sqrt{3}}\)
or\(AB=6\) cm
Hence, the solution of the triangle is
\(\measuredangle C=60^0, BC=2 \sqrt{3} cm, AB=6cm\)
- समकोणी \(\triangle ABC\) मा \(\angle B = 90^\circ\), \(a = \sqrt{3} \text{ cm}\) र \(c = 1 \text{ cm}\) भए, \(\triangle ABC\) को हल गर्नुहोस्। (In a right-angled \(\triangle ABC\), \(\angle B = 90^\circ\), \(a = \sqrt{3} \text{ cm}\) and \(c = 1 \text{ cm}\), solve the \(\triangle ABC\).)
In the given right angled triangle
\(b = \sqrt{a^2 + c^2} = \sqrt{(\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4}\)
or\(AC = 2 \text{ cm}\)
Also
\(\tan A = \frac{p}{b} = \frac{\sqrt{3}}{1} = \sqrt{3}\)
or\(\angle A = 60^\circ\)
Hence,
\(\angle C = 30^\circ\)
Hence, the solution of the triangle is
\(\measuredangle A = 60^\circ,\ \measuredangle C = 30^\circ, AC = 2 cm\)
- \(\triangle MNO\) को हल गर्नुहोस्। जब कि \(\angle M = 90^\circ\), \(\angle N = 30^\circ\) र \(MN = 2 \text{ cm}\) छन्। (Solve the \(\triangle MNO\) where \(\angle M = 90^\circ\), \(\angle N = 30^\circ\) and \(MN = 2 \text{ cm}\).)
In the given right angled triangle
\(\angle O = 60^\circ\)
Also, in \(\triangle MNO\), \(\angle M = 90^\circ\), so \(MN\) and \(MO\) are the legs, and \(NO\) is the hypotenuse.
Given: \(MN = 2 \text{ cm}\) (adjacent to \(\angle N = 30^\circ\))
\(\cos 30^\circ = \frac{MN}{NO} = \frac{2}{NO}\)
or\(\frac{\sqrt{3}}{2} = \frac{2}{NO}\)
or\(NO = \frac{2 \times 2}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3} \text{ cm}\)
Again,
\(\tan 30^\circ = \frac{MO}{MN} = \frac{MO}{2}\)
or\(\frac{1}{\sqrt{3}} = \frac{MO}{2}\)
or\(MO = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \text{ cm}\)
Hence, the solution of the triangle is
\(\measuredangle O = 60^\circ,\ MO = \dfrac{2\sqrt{3}}{3} \text{ cm},\ NO = \dfrac{4\sqrt{3}}{3} \text{ cm}\)
- एउटा \(1.75 \text{ m}\) अग्लो मानिसले एउटा \(51.75 \text{ m}\) उचाई भएको धरहरालाई \(50 \text{ m}\) दुरीबाट अवलोकन गर्दा उन्नतांश कोण कति हुन्छ पत्ता लगाउनुहोस्। (A \(1.75 \text{ m}\) tall man observes the top of a tower \(51.75 \text{ m}\) high from \(50 \text{ m}\) away from the tower. Find the angle of elevation.)
The effective height of the tower above the man's eye level is
\(AE=51.75 \text{ m} - 1.75 \text{ m} = 50 \text{ m}\)
The horizontal distance from the man to the tower CE= \(50 \text{ m}\).
In the right-angled triangle \(\triangle ACE\),
\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{50}{50} = 1\)
or\(\theta = 45^\circ\)
Hence, the angle of elevation is
\(\theta = 45^\circ\)
- एउटी केटीले \(51.5 \text{ m}\) अग्लो धरहरालाई \(50 \text{ m}\) दुरीबाट अवलोकन गर्दा उन्नतांश कोण \(45^\circ\) हुन्छ भने ती केटीको उचाई पत्ता लगाउनुहोस्। (A girl observes the angle of elevation of the top of a tower which is \(51.5 \text{ m}\) high and it is found to be \(45^\circ\). If the distance between the tower and the girl is \(50 \text{ m}\), find the height of the girl.)
Let the height of the girl be \(h\) metres.
The effective height of the tower above the girl's eye level is
\(AE = 51.5 \text{ m} - h\)
The horizontal distance from the girl to the tower is \(CE =BD= 50 \text{ m}\).
In the right-angled triangle \(\triangle ACE\),
\(\tan 45^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{51.5 - h}{50}\)
or\(1 = \frac{51.5 - h}{50}\)
or\(50 = 51.5 - h\)
or\(h = 51.5 - 50 = 1.5\)
Hence, the height of the girl is
\(h = 1.5 \text{ m}\)
- एउटा \(20\sqrt{3} \text{ m}\) अग्लो स्तम्भको टुप्पोमा जमिनको कुनै बिन्दुबाट हेर्दा बनेको उन्नतांश कोण \(60^\circ\) पाइएको भने स्तम्भको फेदबाट उक्त बिन्दुसम्मको दुरी पत्ता लगाउनुहोस्। (The angle of elevation of the top of the tower of \(20\sqrt{3} \text{ m}\) high as observed from a point on the ground is \(60^\circ\); find the distance between the foot of tower and the point on the ground.)
Let the distance between the foot of the tower and the point on the ground be \(x\) metres.
The height of the tower is \(AB = 20\sqrt{3} \text{ m}\).
In the right-angled triangle \(\triangle ABC\), where \(\angle ADB = 60^\circ\),
\(\tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{20\sqrt{3}}{x}\)
or\(\sqrt{3} = \frac{20\sqrt{3}}{x}\)
or\(1 = \frac{20}{x}\)
or\(x = 20\)
Hence, the required distance is
\(x = 20 \text{ m}\)
- एउटा रूख \(15 \text{ m}\) अग्लो छ। जमिनको सतहको कुनै बिन्दुबाट उक्त रूखको टुप्पोमा हेर्दा बनेको उन्नतांश कोण \(45^\circ\) छ भने रूखको फेदबाट उक्त बिन्दुसम्मको दुरी पत्ता लगाउनुहोस्। (The height of a tree is \(15 \text{ m}\). The angle of elevation of the top of the tree as observed from a point on the ground level is \(45^\circ\). Find the distance of the point from the foot of the tree.)
Let the distance between the foot of the tree and the point on the ground be \(x\) metres.
The height of the tree is \(AB = 15 \text{ m}\).
In the right-angled triangle \(\triangle ABC\), where \(\angle ACB = 45^\circ\),
\(\tan 45^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{15}{x}\)
But \(\tan 45^\circ = 1\), so
\(1 = \frac{15}{x}\)
\(x = 15\)
Hence, the required distance is
\(x = 15 \text{ m}\)
- एउटा सिधा स्तम्भको उचाइ \(12 \text{ ft}\) र यसको छायाँको लम्बाइ \(12\sqrt{3} \text{ ft}\) भए, सूर्यको उचाई पत्ता लगाउनुहोस्। (A vertical pole is \(12 \text{ ft}\) high and the length of its shadow is \(12\sqrt{3} \text{ ft}\); find the altitude of the sun.)
Let the altitude (angle of elevation) of the sun be \(\theta\).
The height of the pole is \(AB = 12 \text{ ft}\).
The length of the shadow is \(BC = 12\sqrt{3} \text{ ft}\).
In the right-angled triangle \(\triangle ABC\),
\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{12}{12\sqrt{3}} = \frac{1}{\sqrt{3}}\)
We know that \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so
\(\theta = 30^\circ\)
Hence, the altitude of the sun is
\(\theta = 30^\circ\)
- एउटा स्तम्भको फेदबाट \(50 \text{ m}\) परको दुरीबाट उक्त स्तम्भको टुप्पोमा हेर्दा बनेको उन्नतांश कोण \(45^\circ\) भए, स्तम्भको उचाइ पत्ता लगाउनुहोस्। (The angle of elevation of the top of the tower at a distance of \(50 \text{ meters}\) from the tower is found to be \(45^\circ\). Find the height of the tower.)
Let the height of the tower be \(h\) metres.
The horizontal distance from the observer to the foot of the tower is \(BC = 50 \text{ m}\).
In the right-angled triangle \(\triangle ABC\), where \(\angle ACB = 45^\circ\),
\(\tan 45^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{h}{50}\)
But \(\tan 45^\circ = 1\), so
\(1 = \frac{h}{50}\)
\(h = 50\)
Hence, the height of the tower is
\(h = 50 \text{ m}\)
- एउटा नदीको किनारमा रहेको एउटा \(15 \text{ m}\) अग्लो रूखको टुप्पोमा अर्को किनारबाट हेर्दा बनेको उन्नतांश कोण \(30^\circ\) पाइएको भने नदीको चौडाइ पत्ता लगाउनुहोस्। (A tree on the bank of a river is \(15 \text{ m}\) high and the angle of elevation of the top of the tree from the opposite bank is \(30^\circ\); find the breadth of the river.)
Let the breadth of the river be \(x\) metres.
The height of the tree is \(AB = 15 \text{ m}\).
In the right-angled triangle \(\triangle ABC\), where \(\angle ACB = 30^\circ\),
\(\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{15}{x}\)
But \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so
\(\frac{1}{\sqrt{3}} = \frac{15}{x}\)
Cross-multiplying:
\(x = 15\sqrt{3}\)
Hence, the breadth of the river is
\(x = 15\sqrt{3} \text{ m}\)
- एउटा \(100 \text{ m}\) अग्लो पहाडको टुप्पोबाट जमिनमा रहेको कारलाई हेर्दा अवनति कोण \(30^\circ\) को पाइएको भने पहाडको फेदबाट कारसम्मको दुरी पत्ता लगाउनुहोस्। (From the top of a cliff \(100 \text{ m}\) high, the angle of depression of a car on the ground is found to be \(30^\circ\). Find the distance of the car from the foot of the cliff.)
Let the distance of the car from the foot of the cliff be \(x\) metres.
The height of the cliff is \(AB = 100 \text{ m}\).
The angle of depression from the top of the cliff (\(A\)) to the car (\(C\)) is \(30^\circ\).
Since the horizontal line from \(A\) is parallel to the ground \(BC\), the angle of elevation from \(C\) to \(A\) is also \(30^\circ\).
Thus, in right-angled triangle \(\triangle ABC\),
\(\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{100}{x}\)
But \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so
\(\frac{1}{\sqrt{3}} = \frac{100}{x}\)
Cross-multiplying:
\(x = 100\sqrt{3}\)
Hence, the distance of the car from the foot of the cliff is
\(x = 100\sqrt{3} \text{ m}\)
- एउटा \(30 \text{ ft}\) अग्लो रुख हुरीले भाँचिएर यसको टुप्पोले जमिन छुँदा अधिनतांश \(30^\circ\) को कोण बनाउँछ भने रुखको भाँचिएको भागको लम्बाइ पत्ता लगाउनुहोस्। (A tree of \(30 \text{ ft}\) height is broken by the wind so that its top touches the ground and makes an angle of \(30^\circ\) to the ground. Find the length of the remaining part of the tree after being broken.)
Let the height of the remaining upright part of the tree be \(x\) feet.
Then, the length of the broken part (which now forms the hypotenuse) is \(30 - x\) feet.
The broken top touches the ground at point \(A\), making an angle of \(30^\circ\) with the ground.
In right-angled triangle \(\triangle ABC\) (right angle at \(B\)), the angle at \(A\) is \(30^\circ\).
Here:
– Opposite side to \(30^\circ\) = \(BC = x\)
– Hypotenuse = \(AC = 30 - x\)
Using sine of \(30^\circ\):
\(\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{30 - x}\)
But \(\sin 30^\circ = \frac{1}{2}\), so
\(\frac{1}{2} = \frac{x}{30 - x}\)
Cross-multiplying:
\(30 - x = 2x\)
\(30 = 3x\)
\(x = 10\)
Hence, the length of the remaining (unbroken) part of the tree is
\(x = 10 \text{ ft}\)
- सूर्यको उचाई \(60^\circ\) भएको बेला \(50\sqrt{3} \text{ ft}\) अग्लो स्तम्भको छायाँको लम्बाइ पत्ता लगाउनुहोस्। (Find the length of the shadow of the pole of height \(50\sqrt{3} \text{ ft}\) when the sun's altitude is \(60^\circ\).)
Let the length of the shadow be \(x\) feet.
The height of the pole is \(AB = 50\sqrt{3} \text{ ft}\).
The sun’s altitude (angle of elevation) is \(60^\circ\).
In the right-angled triangle \(\triangle ABC\),
\(\tan 60^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{BC} = \frac{50\sqrt{3}}{x}\)
But \(\tan 60^\circ = \sqrt{3}\), so
\(\sqrt{3} = \frac{50\sqrt{3}}{x}\)
Dividing both sides by \(\sqrt{3}\):
\(1 = \frac{50}{x}\)
\(x = 50\)
Hence, the length of the shadow is
\(x = 50 \text{ ft}\)
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