A) half the sum of the base angles
B) sum of the base angles
C) half the difference of the base angles
D) difference of the base angles
Correct Answer: A
Solution :
As\[\angle BOC={{90}^{o}}-\frac{1}{2}\angle A\] \[\angle A+\angle B+\angle C={{180}^{o}}\] \[\Rightarrow \] \[\frac{1}{2}\angle A={{90}^{o}}-\frac{1}{2}(\angle B+\angle C)\] \[\Rightarrow \] \[\frac{1}{2}(\angle B+\angle C)={{90}^{o}}-\frac{1}{2}\angle A\] Hence, it includes \[\frac{1}{2}\] the sum of the base angles.
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