Question

In $\Delta ABC$, $\angle A={60}^{\circ },\angle B={45}^{\circ }andAB=10cm$ as shown in the figure. Then find the area of the triangle.

Answer 1

In given $\Delta ABC$,

Draw CD perpendicular to AB

$\angle DCB=180-({90}^{\circ }+{45}^{\circ })={45}^{\circ }$

In right angled triangle BDC

$tan{45}^{\circ }=\frac{CD}{DB}=1$

So, $CD=DB=x$

In right angled triangle ADC

$tan{60}^{\circ }=\frac{CD}{AD}$

$\sqrt{3}=\frac{x}{AD}\Rightarrow AD=\frac{x}{\sqrt{3}}$

$AB=AD+DB$

$10=\frac{x}{\sqrt{3}}+x$

$x(\frac{1}{\sqrt{3}}+1)=10$

$x\frac{\sqrt{3}+1}{\sqrt{3}}=10$

$x=\frac{10\sqrt{3}}{\sqrt{3}+1}$

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

=$\frac{1}{2}\times 10\times x$

$=\frac{1}{2}\times 10\times \frac{10\sqrt{3}}{(\sqrt{3}+1)}$

$=\frac{50\sqrt{3}}{\sqrt{3}+1}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}=\frac{50\sqrt{3}(\sqrt{3}-1)}{2}=25(3-\sqrt{3})c{m}^2$