In a (\triangle ABC, ) 2 ac sin ( \bigg \[ \frac{1}{2} (A -... | Mathematics | SolveeIt

Question

In a $\triangle ABC,$ 2 ac sin $\bigg [ \frac{1}{2} (A - B + C ) \bigg ]$ is equal to

$a^2 + b^2 - c^2$

$c^2 + a^2 - b^2$

$b^2 - c^2 - a^2$

$c^2 - a^2 - b^2$

Answer

$c^2 + a^2 - b^2$

Explanation

Solution

We know that, A + B + C = $180^\circ$ $\Rightarrow$ A + C - B = 180 - 2B Now, 2 ac sin $\bigg [ \frac{1}{2} (A - B + C ) \bigg ] = 2ac \, sin \, (90^\circ - B)$ = 2a cos B = $\frac{ 2 ac . (a^2 + c^2 - b^2 }{ 2ac} = a^2 + c^2 - b^2$

Mathematics Question on Three Dimensional Geometry

Solution