# tan (A + B) = √3 and tan (A -B) = 1/√3 | Trigonometry Numerical

## Question: If $\tan (A + B) = \sqrt 3$ and $\tan (A - B) = {1 \over \sqrt 3}$; $0\degree \lt A + B \le 90\degree$; $A \gt B$, find A and B.

Here, we are given that

1. $\tan (A + B) = \sqrt 3$
2. $\tan (A - B) = {1 \over \sqrt 3}$
3. $0\degree \lt A + B \le 90\degree$
4. $A \gt B$

The most important part of this numerical is that we have to remember the values of $tan$ for different angles, as shown in the table below

$\tan 0\degree$ $\tan 30\degree$ $\tan 45\degree$ $\tan 60\degree$ $\tan 90\degree$
$0$ $1 \over \sqrt 3$ $1$ $\sqrt 3$ Not defined

We know from the table above that, $\tan 60\degree = \sqrt 3$. So using this value in the 1st equation from given data:

\begin{aligned} \tan (A + B) &= \sqrt 3 \\ \therefore \tan (A+B) &= \tan 60 \\ \therefore A+B &= 60 --- (5) \end{aligned} Explanation for the above block:
Here we are comparing LHS with RHS. We can say that the equation tan (A+B) equates to √3 and tan 60 is √3 (you can find that in table).
So, this is only possible if A+B = 60 which satisfies this equation.

In the same manner, we know that, $\tan 30\degree = {1 \over \sqrt 3}$. So using this value in the 2nd equation from given data:

\begin{aligned} \tan (A - B) = {1 \over \sqrt 3} \\ \therefore \tan (A-B) &= \tan 30 \\ \therefore A-B &= 30 --- (6) \end{aligned}

Adding equation (5) and equation (6), we get:

\begin{aligned} A + \cancel B = 60 \\ A - \cancel B = 30 \\ \text{\textemdash \textemdash \textemdash \textemdash \textemdash} \\ 2A \qquad = 90 \\ \therefore \boxed {A = 45 \degree} \end{aligned}

Now, we can calculate the value of B by putting the value of A in either equation (5) or equation (6). Let's do it using equation (5):

\begin{aligned} A + B &= 60 \\ \therefore 45 + B &= 60 \\ \therefore \boxed {B = 15 \degree} \end{aligned}

We can further confirm that our answer is correct by looking at equations (3) and equation (4) from the given data that:

\begin{aligned} 0 \lt A &+ B \le 90 \\ \text{And so is } 0 \lt &60 \le 90 \end{aligned}

And,

\begin{aligned} A &\gt B \\ 60 &\gt 15 \end{aligned}

Hence, $\boxed{m\angle A = 60\degree}$ and $\boxed{m\angle B = 15\degree}$