# 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.

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 5

## 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}

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