If sin (A + B + C) = 1, tan (A – B) = \(\frac{1}{\sqrt3}\) and sec (

1.	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» 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°.

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»

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°.