Question

In triangle ABC, $\angle B={30}^{\circ }$ and $\angle C={45}^{\circ }$. If AB =4 cm, then find the length of CD.

• A. 4 cm
• B. 2 cm
• C. $2\sqrt{3}cm$
• D. $4\sqrt{3}cm$

Answer 1

The correct option is B

2 cm

In triangle ABD, $\angle B={30}^{\circ },\angle DAB=180-(90+{30}^{\circ })={60}^{\circ }$

Therefore, angles of triangle ABD 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 corresponding sides will be in the ratio $1:\sqrt{3}:2$.

$\begin{array}{ccccc}{30}^{\circ }& & {60}^{\circ }& & {90}^{\circ }\\ 1& :& \sqrt{3}& :& 2\\ \downarrow & & \downarrow & & \downarrow \\ AD& & BD& & AB\\ \downarrow & & \downarrow & & \downarrow \\ 2cm& & 2\sqrt{3}cm& & 4cm\end{array}$

AD = 2 cm

In $\Delta ADC,\angle DCA={45}^{\circ }$, angle CAD $={180}^{\circ }-({90}^{\circ }+{45}^{\circ })={45}^{\circ }$

Therefore, angles of triangle ADC are ${45}^{\circ },{45}^{\circ },{90}^{\circ }$,

So, the corresponding sides will be in ratio $1:1:\sqrt{2}$.

$\begin{array}{ccccc}{45}^{\circ }& & {45}^{\circ }& & {90}^{\circ }\\ 1& :& 1& :& \sqrt{2}\\ \downarrow & & \downarrow & & \downarrow \\ AD& & DC& & AC\\ \downarrow & & \downarrow & & \downarrow \\ 2cm& & 2cm& & 2\sqrt{2}cm\end{array}$

AD = DC = 2 cm