Ex.8.1 Q4 Introduction to Trigonometry Solution - NCERT Maths Class 10

Go back to  'Ex.8.1'


Given \(15 \,\rm{cot A} = 8,\) find \(\rm{sin\, A}\) and \(\rm{sec\,A.}\)

 Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 4

Text Solution

What is the known?

Cotangent of \(\angle A \)

What is the unknown?

Sine and Secant of \(\angle A \) .


Using cot A, 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 us consider a right-angled \(\Delta \rm{ABC},\) right angled at \(\rm{B.}\)

\[\begin{align}\text{cot}\,\text{A}\, \text{=}\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}\text{=}\frac{\text{AB}}{\text{BC}}\end{align}\]

It is given that

\[\begin{align}\text{cotA}\,\text{=}\,\frac{\text{8}}{\text{15}}\Rightarrow \frac{\text{AB}}{\text{BC}}\,\text{=}\,\frac{\text{8}}{\text{15}}\end{align}\]

Let \(\rm{AB}\) be \(8\,k.\) Therefore, \(\rm{BC}\) will be \(15\,k\) where \(k\) is a positive integer.

Apply Pythagoras theorem in \(\text{ }\!\!\Delta\!\!\text{ }\,\text{ABC,}\) we obtain.

A{C^2}\, &= \,A{B^2} + B{C^2}\\
A{C^2}\, &= \,{(8k)^2} + {(15k)^2}\\
A{C^2}\, &= \,64{k^2} + 225{k^2}\\
A{C^2}\, &= \,289{k^2}\\
\,\;\,AC\,\, &= \,17k

{\rm{sin}}\,{\rm{A}} &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {\rm{A}}}}{{{\rm{hypotenuse}}}} \\ & = \frac{{BC}}{{AC}} = \frac{{15\,k}}{{17\,k}}\\
& = \frac{{{\rm{15}}}}{{{\rm{17}}}}\\
{\rm{sec}}\,{\rm{A}} & = \frac{{{\rm{hypotenuse}}}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle {\rm{A}}}}\\ & = \frac{{AC}}{{AB}} = \frac{{17\,k}}{{8\,k}}\\
 &= \frac{{{\rm{17}}}}{{\rm{8}}}


\[\begin{align}\text{sin}\,\text{A}\,\text{=}\,\frac{\text{15}}{\text{17}}\ \ \text{and}\ \ \text{sec}\,\text{A}\,\text{=}\,\frac{\text{17}}{\text{8}}\end{align}\]