A) \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}\]
B) \[{{c}^{2}}+{{a}^{2}}-{{b}^{2}}\]
C) \[{{b}^{2}}-{{c}^{2}}-{{a}^{2}}\]
D) \[{{c}^{2}}-{{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
We have, \[A+C=\pi -B\] and \[\frac{A-B+C}{2}=\frac{\pi }{2}-B\] \[\therefore \] \[2ca\sin \left( \frac{\pi }{2}-B \right)=\] \[2ca\cos B=2ca\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ca}\] \[={{a}^{2}}+{{c}^{2}}-{{b}^{2}}\]
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