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Question: In a \(\triangle A B C\) \(2 a c \sin \left( \frac { A - B + C } { 2 } \right)\) is equal to...

In a △ABC\triangle A B C 2acsin⁡(A−B+C2)2 a c \sin \left( \frac { A - B + C } { 2 } \right) is equal to

A

a2+b2−c2a ^ { 2 } + b ^ { 2 } - c ^ { 2 }

B

c2+a2−b2c ^ { 2 } + a ^ { 2 } - b ^ { 2 }

C

b2−c2−a2b ^ { 2 } - c ^ { 2 } - a ^ { 2 }

D

c2−a2−b2c ^ { 2 } - a ^ { 2 } - b ^ { 2 }

Answer

c2+a2−b2c ^ { 2 } + a ^ { 2 } - b ^ { 2 }

Explanation

Solution

2acsin⁡A−B+C2=2acsin⁡π−2B2=2accos⁡B2 a c \sin \frac { A - B + C } { 2 } = 2 a c \sin \frac { \pi - 2 B } { 2 } = 2 a c \cos B

=2acc2+a2−b22ca=c2+a2−b22 a c \frac { c ^ { 2 } + a ^ { 2 } - b ^ { 2 } } { 2 c a } = c ^ { 2 } + a ^ { 2 } - b ^ { 2 }.