Question

Please solve this problem

Answer 1

Step-by-step explanation:

Let α, β, γ be the three angles and a, b, c be the side opposite to the angles of a triangle ABC

Relation between α, β, γ

It is given that

3α + 2β = 180º ..........(1)

Sum of three angles of a triangle is 180º

α + β + γ = 180º..........(2)

Substracting (1) and (2) we get

2α + β - γ = 0

or γ = 2α + β

$\sin( \gamma ) = \sin(2 \alpha + \beta )$

$\sin( \gamma ) = \sin(2 \alpha ) \cos( \beta ) + \cos(2 \alpha ) \sin( \beta ) ...</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.</strong><strong>.....(3)$

Sine Formula

$\frac{a}{ \sin( \alpha ) } = \frac{b}{ \sin( \beta ) } = \frac{c}{ \sin( \gamma ) } = k$

$a = k \sin( \alpha )$

$b = k\sin( \beta )$

$c = k \sin( \gamma )$

Proof

LHS =

${a}^{2} + bc$

Using Sine Formula

${k}^{2} \sin {}^{2} ( \alpha ) + {k}^{2} \sin( \beta ) \sin( \gamma )$

Using value of sin(γ) from (3)

${k}^{2} ( \sin {}^{2} ( \alpha ) + \sin( \beta ) ( \sin(2 \alpha ) \cos( \beta ) + \cos(2 \alpha ) \sin( \beta ) )$

${k}^{2} ( \sin {}^{2} ( \alpha ) + \sin(2 \alpha ) \sin( \beta ) \cos( \beta ) + \cos(2 \alpha ) \sin {}^{2} ( \beta ) )$

${k}^{2} ( \sin {}^{2} ( \alpha ) + 2 \sin( \alpha ) \cos( \alpha ) \sin( \beta ) \cos( \beta ) + ( \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) ) \sin {}^{2} ( \beta ) )$

${k}^{2} ( \sin {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) + 2 \sin( \alpha ) \cos( \alpha ) \sin( \beta ) \cos( \beta ) + \cos {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) )$

${k}^{2} ( \sin {}^{2} ( \alpha ) (1 - \sin {}^{2} ( \beta ) ) + 2 \sin( \alpha ) \cos( \alpha ) \sin( \beta ) \cos( \beta ) + \cos {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) )$

${k}^{2} ( \sin {}^{2} ( \alpha ) \cos {}^{2} ( \beta ) + 2 \sin( \alpha ) \cos( \beta ) \cos( \alpha ) \sin( \beta ) + \cos {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) )$

${k}^{2} {( \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta ) )}^{2}$

${k}^{2} \sin {}^{2} ( \alpha + \beta )$

${k}^{2} \sin {}^{2} (\pi - \gamma )$

${k}^{2} \sin {}^{2} ( \gamma )$

$( {k \sin( \gamma )) }^{2}$

${c}^{2}$

= RHS

Hence Proved.