Question

If tan (A+B)=
$\sqrt{3}$

and tan (A-B) =1/
$\sqrt{3}$

;0°<A+B <=90°; A
B, find A and B.​

Answer 1

tan(A+B) = √3

[ tan(π/3) or tan60° = √3 ]

so,

tan(A+B) = tan60°

compare

A+B = 60° ---------equation 1

for second one,

tan(A-B) = 1/√3

[ tan(π/6) or tan30° = 1/√3 ]

so,

tan(A-B) = tan30°

compare

A-B = 30°---------equation 2

Now, solve equation 1 and 2 to get the values of A and B

A + B = 60°

A - B = 30°

----------------

2A - 0 = 90°

2A = 90°

A = 90°/2

A = 45°

put this value in equation 1 or 2 to get value of B

A + B = 60°

45° + B = 60°

B = 60° - 45°

B = 15°