In any - ABC- prove that - bsin B-csin C-asinB-C

To prove b sin B c sin C = a sin (B C) Let sin A/a= sin B/b= sin C/c=1/2R Now a= 2R sin A, b=2R son b amp; c= 2R sin C. ⇒ b sin B c sin C a ...

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Byju's Answer

Standard IX

Mathematics

Adjacent Angles

In any Δ AB...

Question

In any $Δ A B C,$ prove that :- $b sin B - c sin C = a sin (B - C)$

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Solution

To prove
$b s i n B - c s i n C = a s i n (B - C)$
Let
$\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = \frac{1}{2 R}$
Now a= 2R sin A, b=2R son b & c= 2R sin C.
$\Rightarrow b s i n B - c s i n C - a s i n (B - c)$
$= 2 R [s i n^{2} B - s i n^{2} C - s i n (B + C) s i n (B - C)]$ $[∵ s i n A = s i n (180 - A) = s i n (B + C)]$
$R [2 s i n^{2} B - 2 s i n^{2} C - c o s 2 C + c o s 2 B] [∵ 2 s i n x s i n y = c o s (x - y) - c o s (x + y)]$
$\Rightarrow R [2 s i n^{2} B - 2 s i n^{2} C + 1 + 2 s i n^{2} C + 1 - 2 s i n^{2} B]$
$\Rightarrow R \times 0 = 0$
so, b sin B - C sin c- a sin (B-C)=0
b sin B- c sin C = a sin (B-c)

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Similar questions

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

Q. In triangle ABC, prove the following:

b \sin B - c \sin C = a \sin (B - C)

$In any Δ A B C, prove that : a s i n \frac{A}{2} s i n (\frac{B - C}{2}) + b s i n \frac{B}{2} s i n (\frac{C - A}{2}) + c s i n \frac{C}{2} s i n (\frac{A - B}{2}) = 0.$

∣ ∣ ∣ ∣ ∣ \begin{matrix} {sin}^{3} A & {sin}^{3} B & {sin}^{3} C sin A & sin B & sin C cos A & cos B & cos C \end{matrix} ∣ ∣ ∣ ∣ ∣

= - sin (A - B) sin (B - C) sin (C - A) sin (A + B + C)

where

A, B, C

are any three angles such that

A + B + C

\neq n π

Q. In

Δ A B C

prove that

cos (\frac{A - B}{2}) = (\frac{a + b}{c}) sin \frac{c}{2}

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In any ΔABC, prove that :- bsinB−csinC=asin(B−C)

In any $Δ A B C,$ prove that :- $b sin B - c sin C = a sin (B - C)$