Question

1
6. If tan(A+B)= √3 and tan(A-B)= 1/√3 ; 0° <(A+B) ≤ 90°; A ≥ B. Find ‘A’and 'B'.
​

Answer 1

Step-by-step explanation:

RIGHT QUESTION:

tan (A + B) = √3 and tan (A – B) = 1/√3 0° < A + B ≤ 90°; A > B. find A and B.

REQUIRED SOLUTION :

tan (A + B) = √3 ⇒ A + B = 60° ………….. (i)

tan (A – B) = 1/√3 ⇒ A – B = 30° ………… (ii)

From Eq. (i) + Eq.(ii), we have

2A = 90 ⇒ = 45°

A= 45°

From eq. (i), A + B = 60° . .

(i), A + B = 60°

45° + B = 60°

∴ B = 15