Question

The base angles of a triangle are $22{\frac{1}{2}}^0$ and $122{\frac{1}{2}}^0$. Show that the base is equal to twice the height.

Answer 1

$\frac{a}{sinA}=\frac{c}{sin112{\frac{1}{2}}^0}=\frac{c}{cos22{\frac{1}{2}}^0}$ ........(1)
Also $p=csin22{\frac{1}{2}}^0$ .........(2)
$A={180}^0-(22\frac{1}{2}+112{\frac{1}{2}}^0)={45}^0$ ...........(3)
Putting for A and C from (2) and (3) in (1)
$\frac{a}{sin{54}^0}=\frac{p}{sin22{\frac{1}{2}}^{0}cos22{\frac{1}{2}}^0}=\frac{2p}{sin{45}^0}$
$\therefore a=2p$