A) 7 sq. units
B) 6 sq. units
C) 9 sq. units
D) 13 sq. units
Correct Answer: C
Solution :
[c] \[\cos (A-B)=\frac{1-{{\tan }^{2}}\left( \frac{A-B}{2} \right)}{1+{{\tan }^{2}}\left( \frac{A-B}{2} \right)}=\frac{4}{5}\] \[\Rightarrow \,\,\,\tan \left( \frac{A-B}{2} \right)=\frac{1}{3}\] Now, \[\tan \left( \frac{A-B}{2} \right)=\frac{a-b}{a+b}\cot \frac{C}{2}\] \[\Rightarrow \,\,\,\,\cot \frac{C}{2}=1\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,C=\pi /2\] \[\Delta =\frac{1}{2}ab=9\]
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