Question

In the above fIgure. $\angle BAC={60}^0$ and $\angle ABC={65}^0$.

If $DC>DF$, then answer is 1,else 0.

Answer 1

Since sum of angles in a triangle ABC $={180}^o$
$\angle ACB={180}^o-{60}^o-{65}^o={55}^o$
From the figure, $\angle BEC$ is a right angle.
Since sum of angles in a triangle EBC $={180}^o$
$\angle ECB={180}^o-{90}^o-{65}^o={25}^o$
Also, since sum of angles in a triangle FDC $={180}^o$
$\angle DFC={180}^o-{90}^o-{25}^o={65}^o$
Hence, $\angle DFC>\angle ECB$
This means, the sides opposite to the angles will also have the same relation.
$=>DC>DF$