If the angles , and of a triangle are in an arithmetic progression and if , and denote the lengths of the sides opposite to , and respectively, then the value of the expression is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
Since A, B, C are in AP, B = 60°. By the sine rule, a/sin A = c/sin C = 2R, so a = 2R sin A and c = 2R sin C.
The expression becomes:
(a/c)sin 2C + (c/a)sin 2A
= (sin A/sin C)·2 sin C cos C + (sin C/sin A)·2 sin A cos A
= 2 sin A cos C + 2 sin C cos A
= 2 sin(A + C)
= 2 sin(180° − B)
= 2 sin B
= 2 sin 60°
= √3
The expression becomes:
(a/c)sin 2C + (c/a)sin 2A
= (sin A/sin C)·2 sin C cos C + (sin C/sin A)·2 sin A cos A
= 2 sin A cos C + 2 sin C cos A
= 2 sin(A + C)
= 2 sin(180° − B)
= 2 sin B
= 2 sin 60°
= √3
[AIEEE 2010]