Ex.8.2 Q3 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

If \(\begin{align}\tan \,\left( \text{A + B} \right) = \sqrt 3 \end{align}\) and \(\begin{align}\tan \,\left( \rm{A\,-\,B} \right) = \frac{1}{{\sqrt 3 }};\,\,{0^\circ} < \left( \text{A + B} \right) \le 9{0^\circ},\; \rm{A} > \rm{B}, \end{align}\) find \(\text{A}\) and \(\text{B.}\)

Text Solution

Steps:

Given that

\[\begin{align}{\rm{tan}}\;(\rm{A + B}) &= \sqrt 3 \\{\rm{And,}}\;{\rm{tan}}\,(\rm{A} - \rm{B}) &= \frac{1}{{\sqrt 3 }}\\{\rm{Since}}\;\tan {60^\circ} &= \sqrt 3 \;{\rm{and}}\;\tan \,{30^\circ} = \frac{1}{{\sqrt 3 }}\\\text{Therefore,} \\ \therefore {\rm{tan}}\;(\rm{A + B}) &= {\rm{tan}}\;{60^\circ}\\(\rm{A + B}) &= {60^\circ} \qquad \dots{\rm{(i)}}\\ \therefore {\rm{tan}}\;(\rm{A} - \rm{B}) &= \tan \,{30^\circ}\\(\rm{A} - \rm{B}) &= {30^\circ}\qquad \dots{\rm{(ii)}}\end{align}\]

On adding both equations (i) and (ii), we obtain:

\[\begin{align} \rm{A + B + A} - \rm{B} &= {60^\circ} + {30^\circ}\\2\rm{A} &= {90^\circ}\\ \rm{A} &= {45^\circ}\end{align}\]

By substituting the value of \(A\) in equation (i) we obtain

\[\begin{align} \rm{A + B} &= {60^\circ}\\{45^\circ} + \rm{B} &= {60^\circ}\\ \rm{B} &= {60^\circ} - {45^\circ}\\&= {15^\circ}\end{align}\]

Therefore, \(\begin{align}\angle \rm{A} = {45^\circ}\;{\rm{and}}\;\angle \rm{A} = {15^\circ}\;\left( \rm{A > B} \right)\end{align}\)

  
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