Question

In the following figure; angle $B={90}^{\circ },\angle ADB={30}^{\circ }$ and $\angle ACB={60}^{\circ }$.
If $CD=40m$, then $AB$ is 34.64 m
If true then enter $1$ and if false then enter $0$

Answer 1

In $△ABC$
$\tan 60=\frac{AB}{BC}$
$AB=\sqrt{3}BC$
Let $BC=x$, $AB=\sqrt{3}x$
In $△ABD$,
$\tan 30=\frac{AB}{BD}$
$BD=AB\sqrt{3}$
$BC+CD=AB\sqrt{3}$
$x+40=x\sqrt{3}\left(\sqrt{3}\right)$
$x+40=3x$
$x=20$
Thus, $AB=20\sqrt{3}=34.64$ cm