Trigonometric Equations Question 235
Question: The area of triangle $ ABC, $ in which $ a=1,\ b=2 $ , $ \angle C=60{}^\circ $ is
[MP PET 2004]
Options:
A) $ \frac{1}{2} $
B) $ \sqrt{3} $
C) $ \frac{\sqrt{3}}{2} $
D) $ \frac{3}{2} $
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Answer:
Correct Answer: C
Solution:
- We have, $ a=1,,b=2,,\angle C=60{}^\circ $ Area of triangle = $ \frac{1}{2}ab\sin C $ = $ \frac{1}{2}(1),(2),\sin 60{}^\circ =\frac{\sqrt{3}}{2} $ .
BETA
