Question

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

Answer 1

In $△ABC$
$\tan {60}^o=\frac{AB}{BC}$
$\sqrt{3}=\frac{AB}{BC}$
$AB=\sqrt{3}BC$
Let $BC=x$, then $AB=\sqrt{3}x$
In $△ABD$,
$\tan {30}^o=\frac{AB}{BD}$
$\frac{1}{\sqrt{3}}=\frac{AB}{BD}$
$BD=AB\sqrt{3}$
$BC+CD=BD=AB\sqrt{3}$
$x+40=x\sqrt{3}\left(\sqrt{3}\right)$
$x+40=3x$
$x=20$
Thus, $AB=20\sqrt{3}=34.64$ cm
Hence the statement is true and the correct answer to be entered is $1$.