Trigonometric Equations Question 272
Question: In a $ \Delta ABC, $ $ A:B:C $ . Then $
[a+b+c\sqrt{2}] $ is equal to [DCE 2001]
Options:
A) 2b
B) 2c
C) 3b
D) 3a
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \because $ $ A:B:C=3:5:4 $
Þ $ A+B+C=12x=180^{o}\Rightarrow x=15^{o} $
$ \therefore $ $ A=45^{o},,B=75^{o},,C=60^{o} $ $ \frac{a}{\sin 45^{o}}=\frac{b}{\sin 75^{o}}=\frac{c}{\sin 60^{o}}=K $ (say)
$ \therefore ,a=\frac{1}{\sqrt{2}}K,b=\frac{\sqrt{3}+1}{2\sqrt{2}}K,c=\frac{\sqrt{3}}{2}K $
$ \therefore a+b+c\sqrt{2}=3b. $
BETA
