Question

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

Answer 1

Hint: We know that the value of $\tan 60^\circ $ is $\sqrt 3 $ and we also know that the value of $\tan 30^\circ $ is $\dfrac{1}{{\sqrt 3 }}$ . Now, we will put $\tan 60^\circ $ in place of $\sqrt 3 $ and $\tan 30^\circ $ in place of $\dfrac{1}{{\sqrt 3 }}$ and then we will simplify it. In this question we will get two equations and we have to find two unknown variables which we can easily find with the help of the two equations we got.

Complete step-by-step solution:
The information given in the question is $\tan \left( {{\text{A}} + {\text{B}}} \right) = \sqrt 3 $ and $\tan \left( {{\text{A}} - {\text{B}}} \right) = \dfrac{1}{{\sqrt 3 }}$.
We know that the value of $\tan 60^\circ $ is $\sqrt 3 $ and the value of $\tan 30^\circ $ is $\dfrac{1}{{\sqrt 3 }}$
Therefore, we can write $\tan \left( {{\text{A}} + {\text{B}}} \right) = \sqrt 3 $ as:
$
  \tan \left( {{\text{A}} + {\text{B}}} \right) = \sqrt 3 \\
   \Rightarrow \tan \left( {{\text{A}} + {\text{B}}} \right) = \tan 60^\circ
 $
Now, we can write $\left( {{\text{A}} + {\text{B}}} \right) = 60^\circ ……………..(1)$
Similarly, we can write $\tan \left( {{\text{A}} - {\text{B}}} \right) = \dfrac{1}{{\sqrt 3 }}$ as:
$
  \tan \left( {{\text{A}} - {\text{B}}} \right) = \dfrac{1}{{\sqrt 3 }} \\
   \Rightarrow \tan \left( {{\text{A}} - {\text{B}}} \right) = \tan 30^\circ
 $
Hence, we can write $\left( {{\text{A}} - {\text{B}}} \right) = 30^\circ ……………...(2)$
Now, we have two equations and two unknown variables. Therefore we can easily find the value of angle ${\text{A}}$ and ${\text{B}}$ with the help of equation $(1)$ and $(2)$
Now, to find the value of angle ${\text{A}}$ we will add equation $(1)$ and $(2)$ . Therefore, we will get:
$
  \left( {{\text{A}} + {\text{B}}} \right) + \left( {{\text{A}} - {\text{B}}} \right) = 60^\circ + 30^\circ \\
   \Rightarrow 2{\text{A}} = 90^\circ \\
   \Rightarrow {\text{A}} = 45^\circ
 $
Therefore, by adding equation $(1)$ and $(2)$ we got the value of angle ${\text{A}}$
Now, to find the value of angle ${\text{B}}$ we will subtract equation $(2)$ from equation $(1)$ . Therefore, we can write:
$
  \left( {{\text{A}} + {\text{B}}} \right) - \left( {{\text{A}} - {\text{B}}} \right) = 60^\circ - 30^\circ \\
   \Rightarrow 2{\text{B}} = 30^\circ \\
   \Rightarrow {\text{B}} = 15^\circ
 $

Hence, the answer is the value of angle ${\text{A}} = 45^\circ $ and the value of angle ${\text{B}} = 15^\circ $.

Note: The other important things are the formula of $\sin $ , $\cos $ and $\tan $ which we need to memorize.
$\sin {\text{A = }}\dfrac{{{\text{Opposite}}}}{{{\text{Hypotenuse}}}}$
$\cos {\text{A = }}\dfrac{{{\text{Adjacent}}}}{{{\text{Hypotenuse}}}}$
$\tan {\text{A = }}\dfrac{{{\text{Opposite}}}}{{{\text{Adjacent}}}}$
$\cos ec {\text{A = }}\dfrac{{{\text{Hypotenuse}}}}{{{\text{Opposite}}}}$
$\sec {\text{A = }}\dfrac{{{\text{Hypotenuse}}}}{{{\text{Adjacent}}}}$
$\cot {\text{A = }}\dfrac{{{\text{Adjacent}}}}{{{\text{Opposite}}}}$