Triangles And Properties Of Triangle Question 24
Question: If the radius of the cirumcircle of isosceles triangle ABC is equal to AB=AC, then the angle A is:
Options:
A) $ 30{}^\circ $
B) $ 60{}^\circ $
C) $ 90{}^\circ $
D) $ 120{}^\circ $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] 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 isosceles such that $ AB=AC $ Let circumradius be R, then $ R=\frac{AC}{2\sin B}=AB=AC\Rightarrow \frac{AC}{2\sin B}=AC $ $ \sin B=\frac{1}{2}\Rightarrow \sin B=\sin \frac{\pi }{6}\Rightarrow \angle B=\frac{\pi }{6}=\angle C $ We know that $ \angle A+\angle B+\angle C=180{}^\circ =\pi $ $ \angle A+\frac{\pi }{6}+\frac{\pi }{6}=\pi \Rightarrow \angle A+\frac{\pi }{3}=\pi $
$ \Rightarrow \angle A=\pi -\frac{\pi }{3}=\frac{2\pi }{3}=\frac{2\times 180}{3}\Rightarrow \angle A=120{}^\circ $
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