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

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Question: If tan(A+B)=3\tan (A + B) = \sqrt 3 and tan(AB)=13\tan (A - B) = {1 \over \sqrt 3}; 0°<A+B90°0\degree \lt A + B \le 90\degree; A>BA \gt B, find A and B.

Here, we are given that

  1. tan(A+B)=3\tan (A + B) = \sqrt 3
  2. tan(AB)=13\tan (A - B) = {1 \over \sqrt 3}
  3. 0°<A+B90°0\degree \lt A + B \le 90\degree
  4. A>BA \gt B

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

tan0°\tan 0\degree tan30°\tan 30\degree tan45°\tan 45\degree tan60°\tan 60\degree tan90°\tan 90\degree
00 131 \over \sqrt 3 11 3\sqrt 3 Not defined

We know from the table above that, tan60°=3\tan 60\degree = \sqrt 3. So using this value in the 1st equation from given data:

tan(A+B)=3tan(A+B)=tan60A+B=60(5)\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, tan30°=13\tan 30\degree = {1 \over \sqrt 3}. So using this value in the 2nd equation from given data:

tan(AB)=13tan(AB)=tan30AB=30(6)\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:

A+B=60AB=30—————2A=90A=45°\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):

A+B=6045+B=60B=15°\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:

0<A+B90And so is 0<6090\begin{aligned} 0 \lt A &+ B \le 90 \\ \text{And so is } 0 \lt &60 \le 90 \end{aligned}

And,

A>B60>15\begin{aligned} A &\gt B \\ 60 &\gt 15 \end{aligned}

Hence, mA=60°\boxed{m\angle A = 60\degree} and mB=15°\boxed{m\angle B = 15\degree}

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