Question

(b-acosc)sina = acosAsinC

Answer 1

(b - a cosC) sinA = a cosA sinC

Step-by-step explanation:

For any triangles ABC with sides a, b, c assigned to the opposite sides of angles A, B, C, we have the following relations,

a/sinA = b/sinB = c/sinC ..... (1)

cosA = (b² + c² - a²)/(2bc) ..... (2)

cosB = (c² + a² - b²)/(2ca) ..... (3)

cosC = (a² + b² - c²)/(2ab) ..... (4)

From (1), we can write

sinA = ka, sinB = kb, sinC = kc .... (5),

where k is any arbitrary constant

Now, (b - a cosC) sinA

= {b - a (a² + b² - c²)/(2ab)} . ka [by (3), (5)]

= {b - (a² + b² - c²)/(2b)} . ka

= (2b² - a² - b² + c²)/(2b) . ka

= (b² + c² - a²)/(2bc) . kac

= a . (b² + c² - a²)/(2bc) . kc

= a cosA sinC [by (1), (5)]

Hence proved.

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Answer 2

Answer with Step-by-step explanation:

We know that

Sine law:$\frac{Sin A}{a}=\frac{Sin B}{b}=\frac{Sin C}{c}$

$\frac{Sin A}{a}=\frac{Sin B}{b}=\frac{Sin C}{c}=k$

$Sin A=ka, Sin B=kb, Sin C=ck$

Cosine law:

$Cos A=\frac{b^2+c^2-a^2}{2bc}$

$Cos B=\frac{a^2+c^2-b^2}{2ac}$

$Cos C=\frac{a^2+b^2-c^2}{2ab}$

LHS

$(b-aCos C)Sin A$

Substitute the values then we get

$(b-a\times \frac{(a^2+b^2-c^2)}{2ab})\times ka$

$(b-\frac{a^2+b^2-c^2}{2b})ka$

$\frac{(2b^2-a^2-b^2+c^2)}{2bc}\times kac$

$a\times \frac{(b^2+c^2-a^2)}{2bc}(kc)$

Substitute the values then we get

$aCos ASin C$

LHS=RHS

Hence, proved.

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