A) \[{{90}^{o}}-\frac{1}{2}\angle A\]
B) \[{{90}^{o}}+\angle A\]
C) \[{{90}^{o}}+\frac{1}{2}\angle A\]
D) \[{{180}^{o}}-\frac{1}{2}\angle A\]
Correct Answer: C
Solution :
In\[\Delta BOC\], \[\angle 1+\angle 2+\angle BOC={{180}^{o}}\] \[\angle A+\angle B+\angle C={{180}^{o}}\] \[\frac{1}{2}\angle +\frac{1}{2}\angle B+\frac{1}{2}\angle C={{90}^{o}}\] \[\Rightarrow \] \[\frac{1}{2}(\angle A)+\angle 1+\angle 2={{90}^{o}}\] \[\Rightarrow \] \[\angle 1+\angle 2={{90}^{o}}-\frac{1}{2}\angle A\] Put\[\angle 1+\angle 2\]in (i) \[\angle BOC={{180}^{o}}-\left( {{90}^{o}}-\frac{1}{2}\angle A \right)\] \[={{90}^{o}}+\frac{1}{2}\angle A\]You need to login to perform this action.
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