Question

Four point $A,B,C,D$ are in a plane so that $B$ is in the line joining $A$ and $C$ also $B$ is due north of $D$ and $D$ is due west of $C$. $BD=2$, $\angle BDA={45}^{\circ }$ $\angle BCD={30}^{\circ }$. Then $AD$ equal to:

• A. $\sqrt{3}$
• B. $2\sqrt{3}$
• C. $3\sqrt{3}$
• D. None of these

Answer 1

Answer: (D)

None of these

According to Given Conditions we can make the Above Triangle

In $△ABD$

$\frac{AB}{BD}=tan{45}^o\Rightarrow \frac{AB}{2}=1\Rightarrow AB=2$

$A{D}^2=A{B}^2+B{D}^2\Rightarrow AD=\sqrt{2^2+2^2}=2\sqrt{2}$

Therefore, Answer is $2\sqrt{2}$