Question

In the following figure, $\angle BAC={60}^{\circ }$ and $\angle ABC={65}^{\circ }$. Then, which of the following is correct?

1. $CF>AF$

2. $CF<AF$

Answer 1

In this figure triangle ABC

$\angle ACB={180}^o-\angle ABC-BAC$

$\Rightarrow \angle ACB={180}^o-{65}^o-60={55}^o$

In right angle triangle ABD

$\angle BAD={180}^o-\angle ABD-ADB={180}^o-{65}^o-{90}^o={25}^o$

$\angle BAC=\angle BAD+\angle CAD$

$\Rightarrow \angle DAC=\angle BAC-\angle BAD={60}^o-{25}^o={35}^o$

Inright angle triangle AEC

$\angle ACE={180}^o-\angle AEC-BAC\left(orEAC\right)={180}^o-{90}^o-{60}^o={30}^o$

Then in triangle AFC

$\angle DAC($or $FAC)>\angle FE$

$\to {35}^o>{30}^o$

$\to {35}^o>{30}^o$
So, $CF>AF$