Question

In a given $\Delta$ ABC, AB =20 cm, $\angle A={30}^{\circ }$ as shown in figure. If the area of the triangle is $100c{m}^2$, then the length of AC is equal to

• A. $10\sqrt{3}cm$
• B. 10 cm
• C. 20 cm
• D. $20\sqrt{3}cm$

Answer 1

The correct option is C

20 cm

In the given $\Delta$ ABC, AB = 20 cm, $\angle A={30}^{\circ }$

Draw CD perpendicular to AB

Area of traingle $=\frac{1}{2}\times AB\times CD=100$

$\Rightarrow \frac{1}{2}\times 20\times CD=100$

$\Rightarrow CD=10cm$

In $\Delta ADC,\angle ACD={180}^{\circ }-({90}^{\circ }+{30}^{\circ })={60}^{\circ }$

In $\Delta ADC$ angle are ${30}^{\circ },{60}^{\circ },{90}^{\circ }$

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

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

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

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

The corresponding sides are

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ x& :& x\sqrt{3}& :& 2x\\ CD& & AD& & AC\\ \downarrow & & \downarrow & & \downarrow \\ 10& & 10\sqrt{3}& & 20\end{array}$

So AC =20 cm