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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
[tex] \sin( \gamma ) = \sin(2 \alpha + \beta ) [/tex]
[tex] \sin( \gamma ) = \sin(2 \alpha ) \cos( \beta ) + \cos(2 \alpha ) \sin( \beta ) ...................(3)[/tex]
Sine Formula
[tex] \frac{a}{ \sin( \alpha ) } = \frac{b}{ \sin( \beta ) } = \frac{c}{ \sin( \gamma ) } = k[/tex]
[tex]a = k \sin( \alpha ) [/tex]
[tex]b = k\sin( \beta ) [/tex]
[tex]c = k \sin( \gamma ) [/tex]
Proof
LHS =
[tex] {a}^{2} + bc[/tex]
Using Sine Formula
[tex] {k}^{2} \sin {}^{2} ( \alpha ) + {k}^{2} \sin( \beta ) \sin( \gamma ) [/tex]
Using value of sin(γ) from (3)
[tex] {k}^{2} ( \sin {}^{2} ( \alpha ) + \sin( \beta ) ( \sin(2 \alpha ) \cos( \beta ) + \cos(2 \alpha ) \sin( \beta ) )[/tex]
[tex] {k}^{2} ( \sin {}^{2} ( \alpha ) + \sin(2 \alpha ) \sin( \beta ) \cos( \beta ) + \cos(2 \alpha ) \sin {}^{2} ( \beta ) )[/tex]
[tex] {k}^{2} ( \sin {}^{2} ( \alpha ) + 2 \sin( \alpha ) \cos( \alpha ) \sin( \beta ) \cos( \beta ) + ( \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha ) ) \sin {}^{2} ( \beta ) )[/tex]
[tex]{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 ) )[/tex]
[tex]{k}^{2} ( \sin {}^{2} ( \alpha ) (1 - \sin {}^{2} ( \beta ) ) + 2 \sin( \alpha ) \cos( \alpha ) \sin( \beta ) \cos( \beta ) + \cos {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) )[/tex]
[tex] {k}^{2} ( \sin {}^{2} ( \alpha ) \cos {}^{2} ( \beta ) + 2 \sin( \alpha ) \cos( \beta ) \cos( \alpha ) \sin( \beta ) + \cos {}^{2} ( \alpha ) \sin {}^{2} ( \beta ) )[/tex]
[tex] {k}^{2} {( \sin( \alpha ) \cos( \beta ) + \cos( \alpha ) \sin( \beta ) )}^{2} [/tex]
[tex] {k}^{2} \sin {}^{2} ( \alpha + \beta ) [/tex]
[tex] {k}^{2} \sin {}^{2} (\pi - \gamma ) [/tex]
[tex] {k}^{2} \sin {}^{2} ( \gamma ) [/tex]
[tex]( {k \sin( \gamma )) }^{2} [/tex]
[tex] {c}^{2} [/tex]
= RHS
Hence Proved.