Question

In a quadrilateral, $ABCD$ is given that $\angle A={120}^0$, $\angle B=\angle D={90}^0$, $AB=13$, $AD=46$ then $AC$?

• A. $42$
• B. $62$
• C. $72$
• D. $44$

Answer 1

The correct option is B $62$

As given

$\angle A=120°\angle B=90°\angle C=360-(120-90+90)=60°\angle D=90°$

We can see this is cyclic quadrilateral

$AC=2r$ (where $r$ is radius)

$AC=BD/\sin \left(120°\right)=\frac{2}{\sqrt{3}}BD$

By the law of cosine of $\Delta ABD$

$BD\times BD=13\times 13+16\times 46-2\times 13\times 46\cos 120°=2883AC=\frac{2}{\sqrt{3}\sqrt{2883}}=62$