Question

Find the angles of quadrilateral ABCD, in given figure.

Answer 1

AB = BD (in $△$ ADB)
$\therefore \angle 1={70}^{\circ }\therefore \angle 1+{70}^{\circ }+\angle 2={180}^{\circ }\left(Anglessumpropertyoftriangle\right)\Rightarrow 70+70+\angle 2={180}^{\circ }\Rightarrow \angle 2={180}^{\circ }-{140}^{\circ }-{40}^{\circ }Alsoin△DCBDC=CB\therefore \angle 3=\angle 4=y\therefore y+y+80={180}^{\circ }\left(Assumesumpropertyoftriangle\right)\Rightarrow 2y=180-80\Rightarrow y=\frac{100}{2}={50}^{\circ }\therefore \angle 3=\angle 4={50}^{\circ }\therefore \angle ABC=\angle 2+\angle 4={40}^{\circ }+{50}^{\circ }={90}^{\circ }and\angle ADC=\angle 1+\angle 3={70}^{\circ }+{50}^{\circ }={120}^{\circ }\therefore anglesare{70}^{\circ },{90}^{\circ },{80}^{\circ },{120}^{\circ }.$