Question

In $△ABC$ above, the bisector of $\angle BAC$ is perpendicular to $BC$ at point $D$. If $AB=6andBD=3$, what is the measure of $\angle BAC$ ?

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

Answer: (D)

${60}^{\circ }$

Given $AB=6$ and $BD=3$. By hypotenuses theorem we get ${AD}^2={AB}^2-{BD}^2=36-9=27$.

Which gives $AD=3\sqrt{3}$.

Now, if we consider triangle $ABD$, we can see that the ratios of sides of triangle are $1:\sqrt{3}:2$ .

Therefore, angle $BAD$ is equal to ${30}^o$.

Since $AD$ is bisector of angle $BAC$, we get angle $BAC={60}^o$