Question

Right triangle $ABC$ with $\angle A={90}^{\circ }$ has $M$ as the midpoint of $BC$. If $\alpha =2\beta$, then $\frac{AB}{AC}$ equals:

• A. $\frac{1}{2}$
• B. $\sqrt{3}$
• C. $1$
• D. $\frac{\sqrt{3}}{3}$

Answer 1

The correct option is B $\sqrt{3}$

In $△ABC$ and $△AMC$
$\angle BAC=\angle AMC={90}^{\circ }$
$\angle BCA=\angle MCA$ ..... (Common angle)
Thus, $△ABC\sim △MAC$ ........ (By AA rule)
Thus, $\angle B=\angle MAC=\beta$
Now, $\alpha +\beta =90$
$3\beta =90$
$\beta ={30}^{\circ }$
Hence, $\angle B={30}^{\circ }$
Hence, $\cot \angle B=\frac{AB}{AC}$
$\cot {30}^o=\frac{AB}{AC}$
$\frac{AB}{AC}=\sqrt{3}$