Question

In the following figure,$AB=AC$ and ,$AD=AE$ If ,$\angle B={50}^{\circ }$,$\angle D={66}^{\circ }$ and $\angle GAC={18}^{\circ }$ find the measure of AGF

Answer 1

Given: $AB=AC$ and $AD=AE$
In $△ADE$,
$AD=AE$
hence, $\angle ADE=\angle AED=66$ (Isosceles triangle property)
Sum of angles = 180
$\angle ADE+\angle AED+\angle DAE=180$
$66+66+\angle DAE=180$
$\angle DAE=180-132$
$\angle DAE={48}^{\circ }$
In $△ABC$,
$AB=AC$
Hence, by Isosceles triangle property, $\angle ABC=\angle ACB=50$
Sum of angles = 180
$\angle ABC+\angle ACB+\angle BAC=180$
$50+50+\angle BAC=180$
$\angle BAC=80$
$\angle BAF+\angle DAE+\angle GAC=80$
$\angle BAF+48+18=80$
$\angle BAF={14}^{\circ }$
Now, $\angle AGF=\angle GAC+\angle GCA$ (Exterior angle property)
$\angle AGF=18+50$
$\angle AGF={68}^{\circ }$