Trigonometric Equations Question 260
Question: If in a triangle ABC side $ a=(\sqrt{3}+1) $ cms and $ \angle B=30^{o}, $ $ \angle C=45^{o} $ , then the area of the triangle is
[MP PET 1997]
Options:
A) $ \frac{\sqrt{3}+1}{3}cm^{2} $
B) $ \frac{\sqrt{3}+1}{2}cm^{2} $
C) $ \frac{\sqrt{3}+1}{2\sqrt{2}}cm^{2} $
D) $ \frac{\sqrt{3}+1}{3\sqrt{2}}cm^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \Delta =\frac{1}{2}\frac{a^{2}\sin B\sin C}{\sin (B+C)}=\frac{1}{2}\frac{{{(\sqrt{3}+1)}^{2}}.\frac{1}{2}\times \frac{1}{\sqrt{2}}}{\frac{(\sqrt{3}+1)}{2\sqrt{2}}}=\frac{\sqrt{3}+1}{2} $ .
BETA
