Question

In the figure, if$AB\left|\right|CD,EF\left|\right|BC,$, $\angle BAC={65}^{\circ }$ and $\angle DHF={35}^{\circ },$ find $\angle AGH.$

Answer 1

Given : In figure, $AB\left|\right|CD,EF\left|\right|BC,$

$\angle BAC={65}^{\circ },\angle DHF={35}^{\circ }$

$\because EF\left|\right|BC,$

$\therefore \angle A=\angle ACH$ (Alternate angle )

$\therefore \angle ACH={65}^{\circ }$

$\because \angle GHC=\angle DHF$ (Vertivally opposite angles)

$\therefore \angle GHC={35}^{\circ }$

Now in $\Delta GCH,$

Ext. $\angle AGH=\angle GCH+\angle GHC$

=${65}^{\circ }+{35}^{\circ }={100}^{\circ }$