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In a $\triangle ABC, $ 2 ac sin $ \bigg [ \frac{1}{2} (A - B + C ) \bigg ] $ is equal to
In a $\triangle ABC, $ 2 ac sin $ \bigg [ \frac{1}{2} (A - B + C ) \bigg ] $ is equal to
Updated On: Aug 15, 2022
- $ a^2 + b^2 - c^2$
- $ c^2 + a^2 - b^2 $
- $ b^2 - c^2 - a^2$
- $ c^2 - a^2 - b^2 $
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The Correct Option is B
Solution and Explanation
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$
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