Trigonometric Equations Question 315
Question: In a triangle with one angle of $ 120^{o} $ the lengths of the sides form an A. P. If the length of the greatest side is $ 7cm $ , the area of triangle is
Options:
A) $ \frac{3\sqrt{15}}{4}cm^{2} $
B) $ \frac{15\sqrt{3}}{4}cm^{2} $
C) $ \frac{15}{4}cm^{2} $
D) $ \frac{3\sqrt{3}}{4}cm^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- Given that $ 2b=a+c $ and $ c=7cm $ $ \cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}\Rightarrow -\frac{1}{2}=\frac{a^{2}+\frac{a^{2}+c^{2}+2ac}{4}-c^{2}}{2a\frac{(a+c)}{2}} $ On simplification and putting the value of c, we get $ a=3 $ and $ b=5 $ . Hence the area is $ \frac{15\sqrt{3}}{4}cm^{2} $ .
BETA
