A) 7 square unit
B) 8 square unit
C) 9 square unit
D) None of these
Correct Answer: C
Solution :
We have, \[a=6,\,b=3,\,\,\cos (A-B)=\frac{4}{5}\] Let \[t=\tan \left( \frac{A-B}{2} \right)\] \[\cos (A-B)=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\Rightarrow \frac{4}{5}=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}=t=\frac{1}{3}\] So, tan\[\left( \frac{A-B}{2} \right)=\frac{1}{3}\]. Then, \[\tan \left( \frac{A-B}{2} \right)=\frac{a-b}{a+b}\cot \frac{C}{2}\] \[\frac{1}{3}=\frac{6-3}{6+3}\cot \frac{C}{2}\Rightarrow C=90{}^\circ \] Hence, \[\Delta =\frac{1}{2}(6)\,(3)\,\sin 90{}^\circ =9\] square unit.
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