Question: In $ \Delta ABC, $ $ \angle A=30{}^\circ , $ $ \angle B=60{}^\circ . $ Find $ \angle C $ in circular measure.
Options:
A) $ \frac{2{{\pi }^{c}}}{3} $
B) $ \frac{3{{\pi }^{c}}}{4} $
C) $ \frac{{{\pi }^{c}}}{6} $
D) $ \frac{{{\pi }^{c}}}{2} $
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Answer:
Correct Answer: D
Solution:
- In $ \Delta ABC, $
$ \angle A+\angle B+\angle C=180{}^\circ $ [by angle sum property]
$ \angle C=180{}^\circ -(30{}^\circ +60{}^\circ )=90{}^\circ $
$ \because $ $ 180{}^\circ =\pi radian $
$ \therefore $ $ 90{}^\circ =\frac{\pi }{180{}^\circ }\times 90{}^\circ $
$ \therefore $ $ \angle C=\frac{\pi }{2}radian $