In each of the following, one of the six trigonometric ratios is given.

Question:

In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.

(i) $\sin A=\frac{2}{3}$

(ii) $\cos A=\frac{4}{5}$

(iii) $\tan \theta=11$

(iv) $\sin \theta=\frac{11}{15}$

 

(v) $\tan \alpha=\frac{5}{12}$

(vi) $\sin \theta=\frac{\sqrt{3}}{2}$

(vii) $\cos \theta=\frac{7}{25}$

(viii) $\tan \theta=\frac{8}{15}$

(ix) $\cot \theta=\frac{12}{5}$

 

(x) $\sec \theta=\frac{13}{5}$

(xi) $\operatorname{cosec} \theta=\sqrt{10}$

 

(xii) $\cos \theta=\frac{12}{15}$

Solution:

(i) Given: $\sin A=\frac{2}{3}$

By definition,

$\sin A=\frac{\text { Perpendiular }}{\text { Hypotenuse }}$

By Comparing (1) and (2)

We get,

Perpendicular side = 2 and

Hypotenuse = 3

Therefore, by Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

Therefore,

$3^{2}=A B^{2}+2^{2}$

$A B^{2}=3^{2}-2^{2}$

$A B^{2}=5$

$A B=\sqrt{5}$

Hence, Base $=\sqrt{5}$

Now, $\cos A=\frac{\text { Base }}{\text { Hypotenuse }}$

$\cos A=\frac{\sqrt{5}}{3}$

Now, $\operatorname{cosec} A=\frac{1}{\sin A}$

Therefore,

$\operatorname{cosec} A=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} A=\frac{3}{2}$

Now, $\sec A=\frac{\text { Hypotenuse }}{\text { Base }}$

Therefore,

$\sec A=\frac{3}{\sqrt{5}}$

Now, $\tan A=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan A=\frac{2}{\sqrt{5}}$

Now, $\cot A=\frac{\text { Base }}{\text { Perpendicular }}$

Therefore,

$\cot A=\frac{\sqrt{5}}{2}$

(ii) Given: $\cos A=\frac{4}{5}$....(1)

By definition,

$\cos A=\frac{\text { Base }}{\text { Hypotenuse }}$....(2)

By Comparing (1) and (2)

We get,

Base = 4 and

Hypotenuse = 5

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

$5^{2}=4^{2}+B C^{2}$

$B C^{2}=5^{2}-4^{2}$

 

$B C^{2}=25-16$

$B C^{2}=9$

 

$B C=3$

Hence, Perpendicular side = 3

Now, $\sin A=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin A=\frac{3}{5}$

Now, $\operatorname{cosec} A=\frac{1}{\sin A}$

Therefore,

$\therefore \operatorname{cosec} A=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} A=\frac{5}{3}$

Now, $\sec A=\frac{1}{\cos A}$

Therefore,

$\sec A=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec A=\frac{5}{4}$

Now, $\tan A=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan A=\frac{3}{4}$

Now, $\cot A=\frac{1}{\tan \mathrm{A}}$

Therefore,

$\cot A=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot A=\frac{4}{3}$

(iii) Given: $\tan \theta=\frac{11}{1}$....(1)

By definition,

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

By Comparing (1) and (2)

We get,

Base = 1 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

$A C^{2}=1^{2}+11^{2}$

 

$A C^{2}=1+121$

$A C^{2}=122$

$A C=\sqrt{122}$

Hence, Hypotenuse $=\sqrt{122}$

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{11}{\sqrt{122}}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{\sqrt{122}}{11}$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{1}{\sqrt{122}}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{\sqrt{122}}{1}$

$\sec \theta=\sqrt{122}$

Now, $\cot \theta=\frac{1}{\tan \theta}$

Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{1}{11}$

(iv) Given: $\sin \theta=\frac{11}{15}$....(1)

By definition,

$\sin \theta=\frac{\text { Perpendiular }}{\text { Hypotenuse }}$....(2)

By Comparing (1) and (2)

We get,

Perpendicular side = 11 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)

$15^{2}=A B^{2}+11^{2}$

$A B^{2}=15^{2}-11^{2}$

$A B^{2}=225-121$

$A B^{2}=104$

$A B=\sqrt{104}$

$A B=\sqrt{2 \times 2 \times 2 \times 13}$

$A B=2 \sqrt{2 \times 13}$

$A B=2 \sqrt{26}$

Hence, Base $=2 \sqrt{26}$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{2 \sqrt{26}}{15}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{15}{11}$

Now, $\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

Therefore,

$\sec \theta=\frac{15}{2 \sqrt{26}}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{11}{2 \sqrt{26}}$

Now, $\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

Therefore,

$\cot \theta=\frac{2 \sqrt{26}}{11}$

(v) Given: $\tan \alpha=\frac{5}{12}$.....(1)

By definition,

$\tan \alpha=\frac{\text { Perpendicular }}{\text { Base }}$....(2)

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

$A C^{2}=12^{2}+5^{2}$

 

$A C^{2}=144+25$

$A C^{2}=169$

$A C=13$

Hence, Hypotenuse = 13

Now, $\sin \alpha=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \alpha=\frac{5}{13}$

Now, $\operatorname{cosec} \alpha=\frac{1}{\sin \alpha}$

Therefore,

$\operatorname{cosec} \alpha=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \alpha=\frac{13}{5}$

Now, $\cos \alpha=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \alpha=\frac{12}{13}$

Now, $\sec \alpha=\frac{1}{\cos \alpha}$

Therefore,

$\sec \alpha=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \alpha=\frac{13}{12}$

Now, $\cot \alpha=\frac{1}{\tan \alpha}$

Therefore,

$\cot \alpha=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \alpha=\frac{12}{5}$

(vi) Given: $\sin \theta=\frac{\sqrt{3}}{2}$....(1)

By definition,

$\sin \theta=\frac{\text { Perpendiular }}{\text { Hypotenuse }}$....(2)

By Comparing (1) and (2)

We get,

Perpendicular side = and

Hypotenuse = 2

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of perpendicular side (BC) and hypotenuse(AC) and get the base side (AB)

$2^{2}=A B^{2}+(\sqrt{3})^{2}$

$A B^{2}=2^{2}-(\sqrt{3})^{2}$

$A B^{2}=4-3$

$A B^{2}=1$

$A B=\sqrt{1}$

$A B=1$

Hence, Base $=1$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{1}{2}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{2}{\sqrt{3}}$

Now, $\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

Therefore,

$\sec \theta=\frac{2}{1}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{\sqrt{3}}{1}$

Now, $\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

Therefore,

$\cot \theta=\frac{1}{\sqrt{3}}$

(vii) Given: $\cos \theta=\frac{7}{25}$.....(1)

By definition,

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$....(2)

By Comparing (1) and (2)

We get,

Base = 7 and

Hypotenuse = 25

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

$25^{2}=7^{2}+B C^{2}$

$B C^{2}=25^{2}-7^{2}$

$B C^{2}=625-49$

$B C^{2}=576$

$B C=\sqrt{576}$

$B C=24$

Hence, Perpendicular side = 24

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{24}{25}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{25}{24}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{25}{7}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{24}{7}$

Now, $\cot \theta=\frac{1}{\tan \theta}$

Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{7}{24}$

(viii) Given: $\tan \theta=\frac{8}{15}$.....(1)

By definition,

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$....(2)

By Comparing (1) and (2)

We get,

Base = 15 and

Perpendicular side = 8

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

$A C^{2}=15^{2}+8^{2}$

$A C^{2}=225+64$

 

$A C^{2}=289$

$A C=\sqrt{289}$

 

$A C=17$

Hence, Hypotenuse $=17$

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{8}{17}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{17}{8}$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{15}{17}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{17}{15}$

Now, $\cot \theta=\frac{1}{\tan \theta}$

Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{15}{8}$

(ix) Given: $\cot \theta=\frac{12}{5}$....(1)

By definition,

$\cot \theta=\frac{1}{\tan \theta}$.....(2)

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

By Comparing (1) and (2)

We get,

Base = 12 and

Perpendicular side = 5

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)

$A C^{2}=12^{2}+5^{2}$

$A C^{2}=144+25$

$A C^{2}=169$

$A C=\sqrt{169}$

$A C=13$

Hence, Hypotenuse = 13

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{5}{13}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{13}{5}$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{12}{13}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{13}{12}$

Now, $\tan \theta=\frac{1}{\cot \theta}$

Therefore,

$\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

$\tan \theta=\frac{5}{12}$

(x) Given: $\sec \theta=\frac{13}{5}$.....(1)

By definition,

$\sec \theta=\frac{1}{\cos \theta}$.....(2)

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

By Comparing (1) and (2)

We get,

Base = 5 and

Hypotenuse = 13

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

$13^{2}=5^{2}+B C^{2}$

$B C^{2}=13^{2}-5^{2}$

$B C^{2}=169-25$

$B C^{2}=144$

$B C=\sqrt{144}$

$B C=12$

Hence, Perpendicular side = 12

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{12}{13}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{13}{12}$

Now, $\cos \theta=\frac{1}{\sec \theta}$

Therefore,

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

$\cos \theta=\frac{5}{13}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{12}{5}$

Now, $\cot \theta=\frac{1}{\tan \theta}$
Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{5}{12}$

(xi) Given:

$\operatorname{cosec} \theta=\sqrt{10}$....(1)

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

By Comparing (1) and (2)

We get,

Perpendicular side = 1 and

Hypotenuse $=\sqrt{10}$

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)

$(\sqrt{10})^{2}=A B^{2}+1^{2}$

$A B^{2}=(\sqrt{10})^{2}-1^{2}$

$A B^{2}=10-1$

$A B^{2}=9$

$A B=\sqrt{9}$

$A B=3$

Hence, Base side = 3

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{1}{\sqrt{10}}$

Now, $\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$

Therefore,

$\cos \theta=\frac{3}{\sqrt{10}}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{\sqrt{10}}{3}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{1}{3}$

Now, $\cot \theta=\frac{1}{\tan \theta}$

Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{3}{1}$

$\cot \theta=3$

(xii) Given: $\cos \theta=\frac{12}{15}$....(1)

By definition,

$\cos \theta=\frac{\text { Base }}{\text { Hypotenuse }}$.....(2)

By Comparing (1) and (2)

We get,

Base = 12 and

Hypotenuse = 15

Therefore,

By Pythagoras theorem,

$A C^{2}=A B^{2}+B C^{2}$

Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)

$15^{2}=12^{2}+B C^{2}$

$B C^{2}=15^{2}-12^{2}$

$B C^{2}=225-144$

$B C^{2}=81$

$B C=\sqrt{81}$

$B C=9$

Hence, Perpendicular side = 9

Now, $\sin \theta=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$

Therefore,

$\sin \theta=\frac{9}{15}$

Now, $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$

Therefore,

$\operatorname{cosec} \theta=\frac{\text { Hypotenuse }}{\text { Perpendicular }}$

$\operatorname{cosec} \theta=\frac{15}{9}$

Now, $\sec \theta=\frac{1}{\cos \theta}$

Therefore,

$\sec \theta=\frac{\text { Hypotenuse }}{\text { Base }}$

$\sec \theta=\frac{15}{12}$

Now, $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$

Therefore,

$\tan \theta=\frac{9}{12}$

Now, $\cot \theta=\frac{1}{\tan \theta}$

Therefore,

$\cot \theta=\frac{\text { Base }}{\text { Perpendicular }}$

$\cot \theta=\frac{12}{9}$

 

 

 

 

 

 

 

 

 

 

 

 

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