Question

ABCD is a parallelogram. Find the angles $x$, $y$ and $z$ in the given figure.

Answer 1

Since, ABCD is a parallelogram
$\therefore$ ∠ADC = ∠ABC
⇒ $x=z...\left(i\right)$

Now, in right triangle EBC, we have
∠BEC + ∠EBC + ∠ECB = 180°
⇒ 90° + x + 40° = 180°
⇒ x = 180° – 130° = 50°
From $\left(i\right),x=z=$ 50°

Now, in $△FCD$, we have
∠CFD + ∠FDC + ∠FCD = 180°
[Angle sum property]
⇒ 90° + 50° + ∠FCD = 180°
⇒ 140° + ∠FCD = 180°
⇒ ∠FCD = 180° – 140° = 40°

Also, in parallelogram ABCD, we have
∠ADC + ∠DCB = 180°
⇒ 50° + (∠FCD + $y$ + 40°) = 180°
⇒ 50° + 40° + $y$+ 40° = 180°
⇒ $y=$ 180° – 130°
⇒ $y=$50°
$\therefore x=y=z=$ 50°