| Direction: Let ABC be a triangle, R be the circumradius of the triangle. Also, given \[{{R}^{2}}=\frac{1}{8}\,({{a}^{2}}+{{b}^{2}}+{{c}^{2}})\], then |
A) \[\sum{\cos \,2A=-1}\]
B) \[\sum{\cos \,2A=1}\]
C) \[\sum{\sin \,2A=1}\]
D) \[\sum{\sin \,2A=-1}\]
Correct Answer: A
Solution :
By Sine rule, \[\sum{{{a}^{2}}=4{{R}^{2}}\sum{{{\sin }^{2}}A}}\] \[\Rightarrow \] \[4{{R}^{2}}\sum{{{\sin }^{2}}a=8{{R}^{2}}}\] \[(\because \,\,\,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=8{{R}^{2}})\] \[\Rightarrow \] \[2\sum{{{\sin }^{2}}}A=4\] \[\Rightarrow \] \[2\sum{(1-\cos 2A)=4}\] \[\Rightarrow \] \[\sum{\cos \,2A=-1}\]
You need to login to perform this action.
You will be redirected in
3 sec
