With usual notations, in triangle ABC, `a cos (B - C) + b cos (C - A) + c cos (A - B)` is equal to

Question

With usual notations, in triangle ABC, `a cos (B - C) + b cos (C - A) + c cos (A - B)` is equal to

With usual notations, in triangle ABC, `a cos (B - C) + b cos (C - A) + c cos (A - B)` is equal to
A. `(abc)/(R^(2))`
B. `(abc)/(4R^(2))`
C. `(4 abc)/(R^(2))`
D. `(abc)/(2R^(2))`

Monjur Alahi

7 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A
`a cos (B -C) + b cos (C -A) + c co s(A -B)`
`2R sin A co s(B -C) + 2R sin B cos (C - A) + 2R sin C cos (A -B)`
`=2R sin (B+ C) cos (B -C) + 2R sin (A + C) .cos (C -A) + 2R sin (A +B) .cos(A -B)`
`=R [sin 2B + sin 2C + sin 2C + sin 2A + sin 2A + sin 2B]`
`= 2R (sin 2A + sin 2B + sin 2C)`
`=8R sin A sin B sin C`
`=8R (a)/(2R) (b)/(2R)(c)/(2R) = (abc)/(R^(2))`