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# If $$tan (A + B) \ = \ \sqrt{3}$$ and $$tan (A – B) \ = \ \frac{1}{ \sqrt{3}}$$ $$0° < \ A+B \ \le \ 90° \ ; \ A \ > \ B$$ , find A and B.

$$tan (A + B) \ = \ \sqrt{3}$$
=> $$\ tan(A+B) \ = \ tan60°$$
$$\ A+B \ = \ 60°$$ ............(1)

$$tan (A – B) \ = \ \frac{1}{ \sqrt{3}}$$
=> $$\ tan(A-B) \ = \ tan30°$$
$$A – B \ = \ 30°$$...............(2)

Solving (1) and (2), we get

$$\angle$$A = 45° and $$\angle$$B = 15°