Question

In the above figure (not to scale)$\angle BCD={40}^{\circ }$ $\angle EDC={35}^{\circ }$, $\angle CBF={30}^{\circ }$ and $\angle DEG={40}^{\circ }$ find $\angle BAE$

Answer 1

Given $\angle BCD={40}^{\circ }$, $\angle EDC={35}^{\circ }$,$\angle CBF={30}^{\circ }$ and $\angle DEG={40}^{\circ }$
Produce AF to meet CD at P and also produce AG to meet CD at Q.
In $\Delta BPC,\angle BPQ=\angle CBP+\angle BCD$
($\because$ exterior angle) $\Rightarrow \angle BPQ={70}^{\circ }$
In $\Delta EQD,\angle EQP=\angle DEQ+\angle EQDQ$
($\because$ exterior angle)
$\Rightarrow \angle EQP={40}^{\circ }+{35}^{\circ }$
$\Rightarrow \angle EQP={40}^{\circ }+{35}^{\circ }$
In $\Delta APQ,\angle PAQ+\angle APQ+\angle AQP$\
$={180}^{\circ }$
$\Rightarrow \angle PAQ={180}^{\circ }-({70}^{\circ }+{75}^{\circ })={35}^{\circ }.$
$\therefore \angle BAE={35}^{\circ }$