Question
If the radius of the circumcircle of isosceles triangle $$ABC$$ is equal to $$AB = AC,$$ then the angle $$A$$ is :
A.
$${30^ \circ }$$
B.
$${60^ \circ }$$
C.
$${90^ \circ }$$
D.
$${120^ \circ }$$
Answer :
$${120^ \circ }$$
Solution :
If the circumradius of triangle $$ABC$$ be $$R,$$ then
$$R = \frac{a}{{2\sin A}} = \frac{b}{{2\sin B}} = \frac{c}{{2\sin C}}$$
where $$a, b, c$$ has their usual meanings.
Given $$\Delta \,ABC$$ is isoceles such that
$$AB = AC$$
Let circumradius be $$R,$$ then
$$\eqalign{ & R = \frac{{AC}}{{2\sin B}} = AB = AC \cr & \Rightarrow \frac{{AC}}{{2\sin B}} = AC\sin B = \frac{1}{2} \cr & \Rightarrow \sin B = \sin \frac{\pi }{6} \cr & \Rightarrow \angle B = \frac{\pi }{6} = \angle C \cr & {\text{we know that}}\,\angle A + \angle B + \angle C = {180^ \circ } = \pi \cr & \angle A + \frac{\pi }{6} + \frac{\pi }{6} = \pi \cr & \Rightarrow \angle A + \frac{\pi }{3} = \pi \cr & \Rightarrow \angle A = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3} = \frac{{2 \times 180}}{3} \cr & \Rightarrow \angle A = {120^ \circ } \cr} $$
If the circumradius of triangle $$ABC$$ be $$R,$$ then
$$R = \frac{a}{{2\sin A}} = \frac{b}{{2\sin B}} = \frac{c}{{2\sin C}}$$
where $$a, b, c$$ has their usual meanings.
Given $$\Delta \,ABC$$ is isoceles such that
$$AB = AC$$
Let circumradius be $$R,$$ then
$$\eqalign{ & R = \frac{{AC}}{{2\sin B}} = AB = AC \cr & \Rightarrow \frac{{AC}}{{2\sin B}} = AC\sin B = \frac{1}{2} \cr & \Rightarrow \sin B = \sin \frac{\pi }{6} \cr & \Rightarrow \angle B = \frac{\pi }{6} = \angle C \cr & {\text{we know that}}\,\angle A + \angle B + \angle C = {180^ \circ } = \pi \cr & \angle A + \frac{\pi }{6} + \frac{\pi }{6} = \pi \cr & \Rightarrow \angle A + \frac{\pi }{3} = \pi \cr & \Rightarrow \angle A = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3} = \frac{{2 \times 180}}{3} \cr & \Rightarrow \angle A = {120^ \circ } \cr} $$