Question

In triangle ABC,we have $AC=BC=7cm$ and $AB=2cm.$ Suppose that 'D' is a point on the line AB such that B lies between A and D and $CD=8cm$. What is the length of the segment DB?

Answer 1

$S=\frac{7+7+2}{2}=8$

$\therefore area=\sqrt{8(8-7)(8-7)(8-2)}=4\sqrt{3}$

$Also,$

$area=\frac{ab\sin C}{2}=\frac{49\sin C}{2}$

$\therefore \frac{49\sin C}{2}=4\sqrt{3};\sin C=0.2827$

$\therefore \angle C=16.43;\angle A=\angle B={180}^{\circ }-\angle C$

$\therefore \angle A=\angle B=81.7867$

$CE,thealtitudeof△ABC=BC\sin \angle B$

$=7\times \sin 81.7867=6.928cm$

$In△CDE,\sin CDE=\frac{CE}{CD}=\frac{6.928}{8}=0.866$

$\therefore \angle CDE={60}^{\circ }$

$DE=CD\cos 60=8\times 0.5=4cm$

$But,BE=\frac{AB}{2}=\frac{2}{2}=1cm$

$hence,BD=DE-BE=4-1=3cm.$

$\therefore BD=3cm$