Question

In any triangle ABC, prove that :

sinA/a = sinB + sinC / b+c = sinB-sinC / b-c

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Answer 1

In any ∆ABC, using sine rule we say:

a/sinA = b/sinB = c/sinC, where a, b, c are the sides opposite to angle A, B, and C respectively.

Let all these be equal to k.

a/sinA = b/sinB = c/sinC = k

=> a = ksinA ; b = ksinB ; c = ksinC

Then,

• a = ksinA

=> 1/k = sinA/a ... (1)

• b + c = ksinB + ksinC

=> b + c = k(sinB + sinC)

=> 1/k = (sinB + sinC)/(b + c) ...(2)

• b - c = ksinB - ksinC

=> b - c = k(sinB - sinC)

=> 1/k = (sinB - sinC)/(b - c) ...(3)

Therefore, from (1), (2) & (3):

sinA/a=(sinB+sinC)/(b+c)=(sinB-sinC)/(b-c)

Proved.

Answer 2

Given :-

A triangle ABC

To prove :-

sin A/a = sin B + sin C / b + c = sin B-sin C / b-c

Solution :-

Let

$\sf \dfrac{a}{sin \; A} =\dfrac{ b}{sin\; B} = \dfrac{c}{\sin\;C} = x$

Therefore

$\sf a = x \times \sin A$

$\sf a = x sin A$

$\sf b + c = x \; sin B + k\;sin C$

Also

$\sf b -c = x sinB - x sin C$

Solving for b + c

Taking x as common

$\sf k(sin \; B + sin \;C)$

Reciprocal

$\sf \dfrac{1}{x}= \dfrac{ (sinB + sinC)}{(b + c)}$

Solving for b - c

1/b - c =  x(sin B - sin C)

• By cross multiplication

1/x = (sin B - sin C)/(b - c)

From all we concluded

sinA/a = sinB + sinC / b+c = sinB-sinC / b-c