Question

In $\Delta ABC,AB=10cm,\angle A={40}^{\circ },\angle B={80}^{\circ }.$ Find the area of triangle ABC.

$\left[\begin{array}{c}tan{40}^{\circ }=0.83\\ tan{80}^{\circ }=5.6\end{array}\right]$

Answer 1

In $\Delta ABC$ Draw CD perpendicular to AB

Let CD = h, AD = x , so DB becomes 10- x

In triangle CDB

$tan{80}^{\circ }=\frac{CD}{DB}=\frac{h}{(10-x)}$

$h=(10-x).tan{80}^{\circ }$

$h=(10-x).\left(5.6\right)$

$h=56-5.6x...\left(i\right)$

In $\Delta CAD$

$tan{40}^{\circ }=\frac{CD}{AD}=\frac{h}{x}$

$h=xtan{40}^{\circ }$

$h=0.83x...\left(ii\right)$

From (i) $\&$ (ii) we have

$56-5.6x=0.83x$

$56=6.43x$

$x=8.7cm$

From (ii) $h=0.83\times 8.7$

$h=7.22cm$

Area of triangle ABC

$=\frac{1}{2}\times AB\times CD$

$=\frac{1}{2}\times 10\times 7.22=36.14c{m}^2$