Trigonometric Identities Question 347
Question: If $ \sin A=\frac{1}{\sqrt{10}} $ and $ \sin B=\frac{1}{\sqrt{5}}, $ where A and B are positive acute angles, then $ A+B= $
[MP PET 1986]
Options:
A) $ \pi $
B) $ \pi /2 $
C) $ \pi /3 $
D) $ \pi /4 $
Show Answer
Answer:
Correct Answer: D
Solution:
We know that $ \sin ,(A+B)=\sin A\cos B+\cos A\sin B $ $ =\frac{1}{\sqrt{10}}\sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}},\sqrt{1-\frac{1}{10}} $ $ =\frac{1}{\sqrt{10}}\sqrt{\frac{4}{5}}+\frac{1}{\sqrt{5}}\sqrt{\frac{9}{10}}=\frac{1}{\sqrt{50}}(2+3)=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}} $
$ \Rightarrow \sin ,(A+B)=\sin \frac{\pi }{4} $ Hence, $ A+B=\frac{\pi }{4} $ .
BETA
