If sin (A + B + C) = 1, tan (A – B) = 1/√3 and sec (A + C) = 2, then find the values of the angles A, B and C in degrees.

Question

If sin (A + B + C) = 1, tan (A – B) = \(\frac{1}{\sqrt3}\) and sec (A + C) = 2, then find the values of the angles A, B and C in degrees.

Answer 1

sin (A + B + C) = 1 ⇒ sin (A + B + C) = sin 90° ⇒ A + B + C = 90°            …(i) 

tan (A – B) = \(\frac{1}{\sqrt3}\) ⇒ tan (A – B) = tan 30° ⇒ A – B = 30°                   …(ii) 

sec (A + C) = 2 ⇒ sec (A + C) = sec 60° ⇒ A + C = 60°               …(iii) 

Eq (i) – Eq (iii) ⇒ B = 30° 

∴ From eqn (ii), A – 30° = 30° ⇒ A = 60° 

∴ From eqn (i), 60° + 30° + C = 90° ⇒ C = 0° ⇒ A = 60°, B = 30°, C = 0°.