Angles (Isosceles Triangle and Parallelogram)

First

The following figure is not drawn to scale. Given that ABCD is a parallelogram and ∠ADE is an isosceles triangle, find ∠y.

ADC = 180 - 118 = 62 (co-interior angles of parallelogram ABCD)

Y = 62 / 2 (exterior angle = interior opposite angles, base angles of isosceles triangle ADE)

DAB = 118 (diagonally opposite angles of parallelogram ABCD)

y = (180 - 118) / 2 = 31 (co-interior angles of trapezium ABCE)

Second

Study the figure below and find ∠FCB.

Method 1

FCD = EFA = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, corresponding angles)

DCB = 180 - 68 = 112 (co-interior angles of parallelogram ABCD)

FCB = 112 - 48 = 64

Method 2

CFB = (180 - 84) / 2 = 48 (base angles of isosceles triangle EAF, vertically opposite angles)

FBC = 68 (diagonally opposite angles)

FCB = 180 - 48 - 68 = 64 (angle sum of triangle FCB)