Trigonometric Equations Question 293
Question: If in $ \Delta ABC, $ $ a=6,b=3 $ and $ \cos (A-B)=\frac{4}{5} $ , then its area will be
[MP PET 2004]
Options:
A) 7 square unit
B) 8 square unit
C) 9 square unit
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- We have, $ a=6,,b=3,\cos (A-B)=\frac{4}{5} $ Let $ t=\tan ( \frac{A-B}{2} ) $ $ \cos (A-B)=\frac{1-t^{2}}{1+t^{2}}\Rightarrow \frac{4}{5}=\frac{1-t^{2}}{1+t^{2}}=t=\frac{1}{3} $ So, tan $ ( \frac{A-B}{2} )=\frac{1}{3} $ . Then, $ \tan ( \frac{A-B}{2} )=\frac{a-b}{a+b}\cot \frac{C}{2} $ $ \frac{1}{3}=\frac{6-3}{6+3}\cot \frac{C}{2}\Rightarrow C=90{}^\circ $ Hence, $ \Delta =\frac{1}{2}(6),(3),\sin 90{}^\circ =9 $ square unit.
BETA
