Question

If \[\tan \left( {A + B} \right) = \sqrt 3 \] and \[\tan \left( {A - B} \right) = \dfrac{1}{{\sqrt 3 }}\]; \[{0^0} < A + B \leqslant {90^0}\] ; \[A > B\], Find \[A\] and \[B\].

Answer 1

Hint: In this problem, convert the left-hand side values in terms of degrees of tan. Then you will get the two equations in terms of the angles of \[A\],\[B\]. Then solve them accordingly. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

Given \[\tan \left( {A + B} \right) = \sqrt 3 \]

Since \[\tan {60^0} = \sqrt 3 \], we have
\[
   \Rightarrow \tan \left( {A + B} \right) = \tan {60^0} \\
   \Rightarrow \left( {A + B} \right) = {60^0} \\
  \therefore A + B = {60^0}................................................\left( 1 \right) \\
\]

Also given \[\tan \left( {A - B} \right) = \dfrac{1}{{\sqrt 3 }}\]

Since \[\tan {30^0} = \dfrac{1}{{\sqrt 3 }}\], we have
\[
   \Rightarrow \tan \left( {A - B} \right) = \tan {30^0} \\
   \Rightarrow \left( {A - B} \right) = {30^0} \\
  \therefore A - B = {30^0}................................................\left( 2 \right) \\
\]

Adding equation (1) and (2), we get
\[
   \Rightarrow \left( {A + B} \right) + \left( {A - B} \right) = {60^0} + {30^0} \\
   \Rightarrow 2A = {90^0} \\
  \therefore A = {45^0} \\
\]

Subtracting equation (2) from (1), we get
\[
   \Rightarrow \left( {A + B} \right) - \left( {A - B} \right) = {60^0} - {30^0} \\
   \Rightarrow 2B = {30^0} \\
  \therefore B = {15^0} \\
\]

Thus, \[\angle A = {45^0}{\text{ and }}\angle B = {15^0}\].

Note: Here we have fulfilled all the given condition i.e., \[A > B\] as \[\angle A = {45^0}{\text{ and }}\angle B = {15^0}\]. We have taken the values in terms of tan only in the first quadrant \[\left( {{\text{i}}{\text{.e}}{\text{., }}{0^0}{\text{ to }}{{90}^0}} \right)\] because of the given condition \[{0^0} < A + B \leqslant {90^0}\].