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

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Question

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.

Reasoning:

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.

Steps:

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

It is given that:

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

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

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

\[\begin{align} A{C^2} &= A{B^2} + B{C^2}\\
B{C^2} &= A{C^2} - A{B^2}\\
B{C^2} &= {(13\,k)^2} - {(12\,k)^2}\\
B{C^2} &= 169\,{k^2} - 144\,{k^2}\\
B{C^2} &= 25\,{k^2}\\
BC &= 5\,k\end{align}\]

\[\begin{align} \\{\sin \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { hypotenuse }} \\ & =\frac{\mathrm{BC}}{\mathrm{AC}} \\ & =\frac{5}{13} \\ {\cos \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { hypotenuse }} \\ &=\frac{\mathrm{AB}}{\mathrm{AC}} \\ & =\frac{12}{13} \\ {\tan \theta}&=\frac{\text { side opposite to } \angle \theta}{\text { side adjacent to } \angle \theta} \\ &=\frac{\mathrm{BC}}{\mathrm{AB}} \\ & =\frac{5}{12} \\ {\cot \theta}&=\frac{\text { side adjacent to } \angle \theta}{\text { side opposite to } \angle \theta} \\ &=\frac{\mathrm{AB}}{\mathrm{BC}} \\ & =\frac{12}{5}\\ \text{cosec}\,\text{ }\!\!\theta\!\! &=\frac{\text{hypotenuse}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{ }\!\!\theta\!\!\text{ }} \\ & =\frac{\text{AC}}{\text{BC}} \\ & = \frac{\text{13}}{\text{5}}\end{align}\]