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The area of $\Delta$le ABC in which $a=2,b=2.\angle C=60^{\large \circ}$ is

$(a)\;4sq.unit\qquad(b)\;\large\frac{1}{2}$$sq.unit\qquad(c)\;\large\frac{\sqrt 3}{2}$$sq.unit\qquad(d)\;\sqrt 3sq.unit$

1 Answer

We know that Area of a $\Delta$le=$\large\frac{1}{2}$$ab\sin C$
$\qquad\qquad\qquad\qquad\quad\;\;=\large\frac{1}{2}$$\times 2\times 2\times $$\large\frac{\sqrt 3}{2}$
$\qquad\qquad\qquad\qquad\quad\;\;=\sqrt 3$
Hence (d) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
 

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