Trigonometric Equations Question 106
Question: In triangle $ ABC $ , if $ \angle A=45{}^\circ , $ $ \angle B=75{}^\circ $ , then $ a+c\sqrt{2} $ =
[IIT 1988]
Options:
A) 0
B) 1
C) b
D) 2b
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \angle C=180^{o}-45^{o}-75^{o}=60^{o} $ Therefore $ a+c\sqrt{2}=k(\sin A+\sqrt{2}\sin C) $ = $ k,( \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}\sqrt{2} )=k,( \frac{1+\sqrt{3}}{\sqrt{2}} ) $ and $ k=\frac{b}{\sin B}\Rightarrow a+c\sqrt{2}=\frac{b}{\sin 75^{o}}( \frac{1+\sqrt{3}}{\sqrt{2}} )=2b $ .
BETA
