Ex.8.1 Q5 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Given \(\begin{align}\text{sec}\,\theta =\frac{13}{12},\end{align}\) calculate all other trigonometric ratios.

Text Solution

What is the known?

\(\text{Secant}\ \rm{of} \,\theta \)

What is the unknown?

Other trigonometric ratios.


Using \(\rm Sec\;\theta\) , we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.


Let \(\Delta \rm{ABC}\) be a right-angled triangle, right angled at point \(\rm{B.}\)

It is given that:

\[\begin{align}\text{sec}\ \text{ }\!\!\theta\!\!\text{ }\,\text{=}\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }\text{=}\frac{\text{AC}}{\text{AB}}\text{=}\frac{\text{13}}{\text{12}}\end{align}\]

Let \(\rm{AC = 13\,k}\) and \(\rm{BC = 12\,k}\) where \(\rm{k} \) is a positive integer.

Apply Pythagoras theorem in \(\begin{align}\triangle \rm{ABC}\end{align}\) we obtain:

\[\begin{align}  A{C^2} &= A{B^2} + B{C^2}\\
B{C^2} &= A{C^2} - A{B^2}\\
B{C^2} &= {(13\,k)^2} - {(12\,k)^2}\\
B{C^2} &= 169\,{k^2} - 144\,{k^2}\\
B{C^2} &= 25\,{k^2}\\
BC &= 5\,k\\ \\{\sin \theta}&={\frac{\text { side opposite to } \angle \theta}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5}{13}} \\ {\cos \theta}&={\frac{\text { side adjacent to } \angle \theta}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}} \\ {\tan \theta}&={\frac{\text { side opposite to } \angle \theta}{\text { side adjacent to } \angle \theta}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}} \\ {\cot \theta}&={\frac{\text { side adjacent to } \angle \theta}{\text { side opposite to } \angle \theta}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12}{5}}\\ \text{cosec}\,\text{ }\!\!\theta\!\! &=\frac{\text{hypotenuse}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{ }\!\!\theta\!\!\text{ }}\text{=}\frac{\text{AC}}{\text{BC}}\text{=}\frac{\text{13}}{\text{5}}\end{align}\]