Question

If $tan(A-B)=\frac{1}{√3}$ and tan (A+B)=√3, 0° < A+B≤90°, A>B find A and B.

Answer 1

SOLUTION :

Given:

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

tan (A - B) = 1/√3

tan (A - B) = tan 30°

[tan 30° = 1/√3]

(A - B) =  30° ………..(1)

tan (A + B) = √3

tan (A + B) = tan 60°

[tan 60° = √3]

(A + B) =  60° ………..(2)

On adding eq 1 & 2,  

A + B = 60°

A -  B = 30°

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2A   = 90°

A = 90/2 = 45

A = 45°

On putting the value of  A = 45° in eq 1

A -  B = 30°

45° - B = 30°

B = 45° - 30° = 15°

B = 15°

Hence, the value of A = 45° and B = 15°

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

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