Question

In a pentagon ABCDE, AB is parallel to DC and $\angle A:\angle E:\angle D=3:4:5.$ Find angle E.

Answer 1

In a pentagon ABCDE, AB is parallel to DC and $\angle A:\angle E:\angle D=3:4:5.So,\angle B+\angle C={180}^{\circ }$
(Sum of interior angles n the same side of transversal are supplementary.)
$Let\angle A=3x,\angle E=4x,\angle D=5x$
Sum of all interior angles of pentagon $={180}^{\circ }(n-2)={180}^{\circ }(5-2)={540}^{\circ }$
Now,
$\angle B+\angle C+\angle A+\angle E+\angle D={540}^{\circ }{180}^{\circ }+3x+4x+5x={540}^{\circ }12x={360}^{\circ }x={30}^{\circ }\angle E=4c=4\times {30}^{\circ }={120}^{\circ }$