Question

Find the area of the isosceles triangle with base $16cm$ and vertical angle ${60}^{\circ }{40}^{\prime }$

Answer 1

Let ${}^{\prime }{h}^{\prime }$be the height of the isosceles triangle and $BC$ be the base.

$\Rightarrow \tan 60°{40}^{\prime }=\tan 60.67°$

Now, In $△ADC$

$\Rightarrow \tan 60.67°=\frac{AD}{DC}=\frac{h}{8}$

$\Rightarrow h=8\tan 60.67°=8\times 1.7798=14.2384cm$

Now, $area△ABC=\frac{1}{2}\times BC\times h$

$=\frac{1}{2}\times 16\times 14.2384$

$=113.9072{cm}^2$

Hence, the answer is $113.9072{cm}^2.$