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