Ex.8.1 Q5 Introduction to Trigonometry Solution - NCERT Maths Class 10
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:
\[\begin{align}\text{sec}\ \text{ }\!\!\theta\!\!\text{ }\,\text{=}\frac{\text{hypotenuse}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \theta }\text{=}\frac{\text{AC}}{\text{AB}}\text{=}\frac{\text{13}}{\text{12}}\end{align}\]
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:
\[\begin{align} A{C^2} &= A{B^2} + B{C^2}\\
B{C^2} &= A{C^2} - A{B^2}\\
B{C^2} &= {(13\,k)^2} - {(12\,k)^2}\\
B{C^2} &= 169\,{k^2} - 144\,{k^2}\\
B{C^2} &= 25\,{k^2}\\
BC &= 5\,k\\ \\{\sin \theta}&={\frac{\text { side opposite to } \angle \theta}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{5}{13}} \\ {\cos \theta}&={\frac{\text { side adjacent to } \angle \theta}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}} \\ {\tan \theta}&={\frac{\text { side opposite to } \angle \theta}{\text { side adjacent to } \angle \theta}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{5}{12}} \\ {\cot \theta}&={\frac{\text { side adjacent to } \angle \theta}{\text { side opposite to } \angle \theta}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12}{5}}\\ \text{cosec}\,\text{ }\!\!\theta\!\! &=\frac{\text{hypotenuse}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{ }\!\!\theta\!\!\text{ }}\text{=}\frac{\text{AC}}{\text{BC}}\text{=}\frac{\text{13}}{\text{5}}\end{align}\]