Question

In Figure 3, $△$ $ABC$ is equilateral and $△$ $ADC$ is isosceles. If $AC=1$, calculate the distance from $B$ to $D$.

• A. $0.286$
• B. $0.318$
• C. $0.333$
• D. $0.366$
• E. $0.383$

Answer 1

The correct option is D $0.366$

Given, $AC=1$ which implies

$AD=DC=\frac{1}{\sqrt{2}}$ and $AB=BC=1$
$\angle BAD=15$ degrees and $\angle ABD=30$ degrees.
By sine rule, we get
$\Rightarrow \frac{BD}{\sin 15}=\frac{AD}{\sin 30}=\frac{1}{\sqrt{2}\times \frac{1}{2}}=\sqrt{2}$
$\Rightarrow BD=\sqrt{2}\times \sin 15=0.366$