# Ex.8.1 Q3 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

If \begin{align}\text{sin A}=\frac{3}{4}\end{align} calculate $$\rm{cos\,A}$$ and $$\rm {tan\,A}.$$

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 3

## Text Solution

#### What is the known?

Sine of $$\angle \text{A}$$ .

#### What is the unknown?

Cosine and tangent of $$\angle \text{A}$$

#### Reasoning:

Using sin 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 $$\Delta \text{ABC}$$ be a right-angled triangle, right angled at point $$\rm{B}$$.

Given that:

\begin{align} \Rightarrow {\sin A}&={\frac{3}{4}} \\ {\frac{B C}{A C}}&={\frac{3}{4}}\end{align}

Let $$\rm{BC}$$ be $$3k$$. Therefore, $$\rm{AC}$$ will be $$4k$$ where $$k$$ is a positive integer.

Applying Pythagoras theorem for $$\Delta \,\rm ABC,$$  we obtain:

\begin{align} A{C^2} &= A{B^2}\, + \,B{C^2}\\ A{B^2} &= A{C^2} - \,B{C^2}\\ A{B^{2\,}} &= {(4\,k)^2} - \,{(3\,k)^2}\\ A{B^2} &= 16{k^2} - 9\,{k^2}\\ A{B^2}\, &= \,7\,{k^2}\\ AB\, &= \sqrt {7\,} k \end{align}

\begin{align} \,\text{cosA}\,&=\frac{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}}{\text{hypotenuse}} \\ & =\frac{\text{AB}}{\text{AC}}\\ & =\frac{\sqrt{\text{7}}k}{\text{4}\,k} \\&=\frac{\sqrt{\text{7}}}{\text{4}} \end{align}

\begin{align} \text{tan}\,\text{A}&=\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{A}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{A}} \\ & =\frac{\text{BC}}{\text{AB}}\,\,\,\text{=}\frac{\text{3}\,k}{\sqrt{\text{7}}\,k} \\ &=\frac{\text{3}}{\sqrt{\text{7}}} \end{align}

Thus,

\begin{align}\,\text{cos}\,\text{A=}\frac{\sqrt{\text{7}}}{\text{4}}\ \ \text{and}\ \text{tan}\ \text{A=}\frac{\text{3}}{\sqrt{\text{7}}}\end{align}

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