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Let $\triangle ABC$ and $\triangle ABC'$ be two non congruent triangles with side $AB=4$, $AC=AC'=2$$\sqrt{2}$ and $\displaystyle\angle B=30^\circ$. Find the absolute value of the difference between the area of these triangles.

Diagram:- http://prntscr.com/1vl225

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    $\begingroup$ by $\angle B$ you mean $\angle ABC,$ or $\angle ABC'?$ $\endgroup$ – lab bhattacharjee Oct 6 '13 at 9:35
  • $\begingroup$ both will be same $\endgroup$ – dknight Oct 6 '13 at 9:35
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HINT:

Apply $$\cos 30^\circ=\frac{a^2+4^2-(2\sqrt2)^2}{2\cdot a\cdot 4}$$ where the values of $a$ represent $BC,BC'$

The area of $\triangle ABC=\frac12BC\cdot AB\cdot\sin30^\circ$ and similarly for $\triangle ABC'$

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Hint: Let $D$ be the midpoint of $CC'$. Then $AD \perp BC'$ and hence $AD=AB\sin \angle B$.

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