In an acute–angled triangle ABC if sin (B + C

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In an acute–angled triangle ABC if sin (B + C – A) = √3 and tan (C + A – B) = 1, then C is equal to
2

1. 37.5°
2. 67.5°
3. 52.5°
4. 72.5°

Correct Option: C

As per the given question ,

Sin (B + C – A) =	√3	= sin 60°
	2

⇒ B + C – A = 60° ..... (i)
Again,
tan (C + A – B) = 1 = tan 45°
⇒ C + A – B = 45° ..... (ii)
On adding (i) and (ii) , we get
∴ B + C – A + C + A – B = 60° + 45°
⇒ 2C = 105°

⇒ C =	105°	= 52.5°
	2

In an acute–angled triangle ABC if sin (B + C – A) =	√3	and tan (C + A – B) = 1, then C is equal to
	2