If sin(A + B + C) = 1, then tan(A – B) = 1/√3 and sec(A + C) = 2, find A, B and C respectively when they are acute.
(A) 60°, 0°, 30°
(B) 30°, 60°, 90°
(C) 60°, 30°, 0°
(D) 0°,60°, 30°
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If sin(A + B + C) = 1, then tan(A – B) = 1/√3 and sec(A + C) = 2, find A, B and C respectively when they are acute.
(A) 60°, 0°, 30°
(B) 30°, 60°, 90°
(C) 60°, 30°, 0°
(D) 0°,60°, 30°
If sin(A + B + C) = 1, then tan(A – B) = 1/√3 and sec(A + C) = 2, find A, B and C respectively when they are acute.
(A) 60°, 0°, 30°
(B) 30°, 60°, 90°
(C) 60°, 30°, 0°
(D) 0°,60°, 30°
The correct option is: (C) 60°, 30°, 0°
Explanation:
We have, sin(A + B + C) = 1
=> sin(A + B + C) = sin 90°
=> A + B + C = 90° …(i)
Also, tan(A – B) = 1/√3 = tan 30°
=> A – B = 30° …(ii)
and sec (A + C) = 2 = sec 60°
=> A + C = 60° ….(iii)
From (ii) and (iii), we get
B + C = 30° …(iv)
From (i) and (iv), we get, A = 60°
. .. B = 30° [Using A = 60° in (ii)]
and C = 0° [Using A = 60° in (iii)]