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$ABC$ is a $\Delta$le.Its area is 12sqcm and base is 6cm.The difference of base angle is $60^{\large\circ}$.If $A$ be the angle opposite to the base,then the value of $8\sin A-6\cos A$ is


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Given :
Area $\Delta=12cm^2$
Now $\Delta=\large\frac{1}{2}$$ab\sin C$
$\qquad\;\;\;=\large\frac{1}{2}$$a.k\sin B\sin C$
$\qquad\;\;\;=\large\frac{1}{2}$$a.\large\frac{a}{\sin A}$$\sin B\sin C$
$\large\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$=k$
$\Delta=\large\frac{1}{2}\frac{a^2\sin B\sin C}{\sin A}$
$12=\large\frac{1}{2}\frac{36\sin B\sin C}{\sin A}$
$\large\frac{2}{3}=\frac{2\sin B\sin C}{2\sin A}$
$\large\frac{2}{3}=\frac{\cos(B-C)-\cos(B+C)}{2\sin A}$
$\large\frac{2}{3}=\frac{\cos (60^{\large\circ})-\cos(\pi-A)}{2\sin A}$
$\large\frac{2}{3}=\frac{1+2\cos A}{4\sin A}$
$4\sin A=3+6\cos A$
$8\sin A-6\cos A=3$
Hence (c) is the correct answer.
answered Nov 18, 2013 by sreemathi.v
edited Mar 5, 2014 by meenakshi.p

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