Question

In a triangle ABC, AB =20 cm, $\angle A={30}^{\circ }$, AC = 18 cm. Find the area of the triangle

Answer 1

Draw CD perpendicular to AB.

In triangle ADC
$\angle C={30}^{\circ }.....\left(Anglesumproperty\right)$
Hence, all the angles of triangle ADC are :
${30}^{\circ },{60}^{\circ },{90}^{\circ }$
Now,
$Theratiosofthesineofangles=\sin \left(30\right):\sin \left(60\right):\sin \left(90\right)$

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

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

So, the sides CD, AD, AC will be in the ratio $1:\sqrt{3}:2$, therefore the corresponding sides can be calculated as

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ CD& & AD& & AC\\ \downarrow & & \downarrow & & \downarrow \\ 9& & 9\sqrt{3}& & 18\end{array}$

So, the perpendicular CD is equal to 9 cm

Area of triangle $=\frac{1}{2}\times AB\times CD=\frac{1}{2}\times 20\times 9$

$=90c{m}^2$