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 \(\rm{sec\,A.}\)

Text Solution

What is the known?

Cotangent of \(\angle \text{A}\)

What is the unknown?

Sine and Secant of \(\angle \text{A}\) .

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

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

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

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

Thus,\(\begin{align}\text{sin}\,\text{A}\,\text{=}\,\frac{\text{15}}{\text{17}}\ \ \text{and}\ \ \text{sec}\,\text{A}\,\text{=}\,\frac{\text{17}}{\text{8}}\end{align}\)