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

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## Question

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 $$\angle A$$ .

#### Reasoning:

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.

#### Steps:

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.

\begin{align} 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 \end{align}

\begin{align} {\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}}} \end{align}

Thus,

\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}

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