Question

In $△\text{ABC}$, AB=8 cm, $\angle A={45}^{\circ }$, $\angle B={60}^{\circ }$.
Find the area of the triangle.

Answer 1

Let AD = x , then DB = (8-x) cm

In $△$ADC, angles are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$ the corresponding sides can be calculated as

$\Rightarrow \sin \left(45\right):\sin \left(45\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{\sqrt{2}}:\frac{1}{\sqrt{2}}:1$

$\Rightarrow 1:1:\sqrt{2}$

${45}^{\circ }{45}^{\circ }{90}^{\circ }1:1:\sqrt{2}\downarrow \downarrow \downarrow ADCDAC\downarrow \downarrow \downarrow xx\sqrt{2x}$

In $△$DBC, angles are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$ the corresponding sides can be calculated as

$\Rightarrow \sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

$\Rightarrow \frac{1}{2}:\frac{\sqrt{3}}{2}:1$

$\Rightarrow 1:\sqrt{3}:2$

${30}^{\circ }{60}^{\circ }{90}^{\circ }1:\sqrt{3}:2\downarrow \downarrow \downarrow DBCDCB\downarrow \downarrow \downarrow \frac{x}{\sqrt{3}}x\frac{2x}{\sqrt{3}}$

Now, AB = $AD+DB=8(x+\frac{x}{\sqrt{3}})=8x(1+\frac{1}{\sqrt{3}})=8x\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)=8x=\frac{8\sqrt{3}}{(\sqrt{3}+1)}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}=\frac{8\sqrt{3}(\sqrt{3}-1)}{2}=4\sqrt{3}(\sqrt{3}-1)$

Area of triangle = $\frac{1}{2}\times AB\times CD=\frac{1}{2}\times 8\times x=\frac{1}{2}\times 8\times 4\sqrt{3}(\sqrt{3}-1)$

Hence, area = $16\sqrt{3}(\sqrt{3}-1)c{m}^2$

= $48-16\sqrt{3}c{m}^2$