Question

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

Answer 1

Answer:

A= 45° and B= 15°

Step-by-step explanation:

tan(A-B)= 1/√3

(A-B)=30°.....(1)

tan(A+B)=√3

(A+B)=60°....(2)

From (1) &(2),

A=45° and B= 15°

Answer 2

A = 45°; B = 15°

Step-by-step explanation:

Given: tan (A – B) =1/√3 and tan (A + B) =√3 , 0° < A + B ≤90°, A > B

Find: A and B

Solution:  

 tan (A – B) = 1/√3

 tan 30° = 1/√3

 So we get A - B = 30° -------(1)

 tan (A + B) = √3

 tan 60° = 1

 So we get A + B = 60° --------(2)

 Adding (1) & (2) = 2A = 90°

 Therefore A = 45°.

 Substituting in (1), we get 45° - B = 30°  

 Therefore B = 15°.