A) \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\]
B) \[{{b}^{2}}+{{c}^{2}}={{a}^{2}}\]
C) \[{{c}^{2}}+{{a}^{2}}={{b}^{2}}\]
D) \[b-c=c-a\]
Correct Answer: A
Solution :
\[\cos \frac{A}{2}=\sqrt{\frac{b+c}{2c}}\Rightarrow \sqrt{\frac{b+c}{2c}}=\sqrt{\frac{s(s-a)}{bc}}\] or \[{{b}^{2}}+bc=2s(s-a)\] \[\Rightarrow \] \[{{b}^{2}}+bc=(a+b+c)\text{ }\left( \frac{b+c-a}{2} \right)\] \[\Rightarrow \]\[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\].
You need to login to perform this action.
You will be redirected in
3 sec
