A) \[5\sqrt{2}\]
B) 9
C) \[6\sqrt{2}\]
D) 11
E) \[13\]
Correct Answer: B
Solution :
Let \[a=6,b=5,c=\sqrt{13}\] \[\therefore \] \[\cos C=\frac{{{6}^{2}}+{{5}^{2}}-13}{2\times 6\times 5}=\frac{48}{2\times 6\times 5}\] \[=\frac{4}{5}\] Now, \[\sin C=\sqrt{1-\frac{16}{25}}=\frac{3}{5}\] \[\therefore \]Area of \[\Delta ABC=\frac{1}{2}ab\,\sin C\] \[=\frac{1}{2}\times 6\times 5\times \frac{3}{5}=9\,sq\,unit\]
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