For the triangle ABC, if B or C is an obtuse angle, then a = b cos C + c cos B according to the law of sines and the addition formula.
For a triangle ABC with angles ∠A, ∠B, and ∠C and with sides a, b, and c
Thus, we can write,
sinA/a = sinB/b = sinC/c
or
a/sinA = b/sinB = c/sinC
It is given that in a ΔABC,
The ∠B or ∠C is an obtuse angle.
So, ∠A = 180° - (B + C)
⇒ sin A = sin (180° - (B + C))
⇒ sin A = sin (B + C) (since we know that sin(180° - θ) = sinθ)
From the addition formula,
sin (B + C) = sin B cos C + sin C cos B
⇒ sin A = sin (B + C)
⇒ sin A = sin B cos C + sin C cos B
From the law of sines,
a/sinA = b/sinB = c/sinC = K
⇒ aK =sin A, bK = sin B, cK = sin C
On substituting,
⇒ sin A = sin B cos C + sin C cos B
⇒ aK = bK cos C + cK cos B
⇒ K(a) = K(b cos C + c cos B)
∴ a = b cos C + c cos B
Hence it is proved.
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