Question

If tan (A+B)=
$\sqrt{3}$
and tan (A-B) =
$1 \div \sqrt{3}$
0° < A+B <90° A>B find
A and B.

​

Answer 1

Answer:

tan(A+B) = √3 = tan60°

A+B = 60° ...........(1)

tan(A-B) = 1/√3 = tan30°

A-B = 30° ............(2)

from (1) and (2)

2A = 90°

A= 45°

from (1)

B = 15°