In any - ABC- prove that-a sin B -sin C - b sin C -sin A - c sin A -sin B -0

By the sine rule, we have a/sin A = b/sin B = c/sin C = K (say) ⇒ a = k sin A, b = k sin B and c = k sin C ∴ LHS = a(sin B sin C) + b(sin C sin A) + c(si ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Quadratic Formula

In any Δ ABC,...

Question

In any $Δ A B C$ , prove that.

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$

Open in App

Solution

By the sine rule, we have

$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = K (s a y)$

$\Rightarrow a = k s i n A, b = k s i n B a n d c = k s i n C$

$∴ L H S = a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B)$

= $k s i n A (s i n B - s i n C) + k s i n B (s i n C - s i n A) + k s i n C (s i n A - s i n B)$

= $K [(s i n A s i n B - s i n A s i n C) + (s i n B s i n C - s i n A s i n B) + (s i n A s i n C - s i n B s i n C)]$

$(k \times 0) = 0 = R H S .$

Suggest Corrections

Similar questions

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$

Q. In triangle ABC, prove the following:

a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0

Q. In

Δ A B C

, a

s i n (B - C) + b s i n (C - A) + c s i n (A - B) =

Q. In

△ A B C

, prove that a sin (B - C) + b sin (C - A) + c sin (A - B) = 0

Q. In any

Δ A B C

s i n A + s i n B + s i n C =

Join BYJU'S Learning Program

In any ΔABC, prove that. a(sin B−sin C)+b (sin C−sin A)+c (sin A−sin B)=0

In any $Δ A B C$ , prove that.

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$