Trigonometric Equations Question 87
Question: If in a triangle $ ABC, $ $ \cos A\cos B+\sin A\sin B\sin C=1, $ then the sides are proportional to
Options:
A) 1: 1: $ \sqrt{2} $
B) $ 1:\sqrt{2}:1 $
C) $ \sqrt{2}:1:1 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- From the given relation $ \sin C=\frac{1-\cos A\cos B}{\sin A\sin B}\le 1 $ ?..(i)
Þ $ 1\le \cos A\cos B+\sin A\sin B $
Þ $ \cos (A-B)\ge 1 $ ; $ \because \cos \theta \not{>}1 $ ?..(ii) \ $ A-B=0 $ or $ A=B $ Hence from (i), $ \sin C=\frac{1-{{\cos }^{2}}A}{{{\sin }^{2}}A}=\frac{{{\sin }^{2}}A}{{{\sin }^{2}}A}=1 $ \ $ C=90^{o}\Rightarrow A+B=90^{o} $ or $ A=B=45^{o} $ {by (ii)} Hence, $ a:b:c $ = $ \sin A:\sin B:\sin C=1:1:\sqrt{2} $ .
BETA
