Question

In the above fIgure. $\angle BAC={60}^0$ and $\angle ABC={65}^0$. Can we say $CF>AF$?

If true enter 1 or else enter 0.

Answer 1

Since sum of angles in a triangle $={180}^o$
$\angle ACB={180}^o-{60}^o-{65}^o={55}^o$
From the figure, $\angle FDC>\angle ACF$ since $\angle FDC$ is right angle and $\angle ACF$ is an acute angle.

This means, the sides opposite to the angles will also have the same relation.
$=>CF>AF$