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 \(\rm {tan\,A}.\)

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 \(\rm{3k}\). Therefore, \(\rm{AC}\) will be \(\rm{4k}\) where \(\rm{k} \) is a positive integer.

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

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

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

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

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