Question

In the given figure, find $\angle ADG$ and $\angle IEJ$. ABCD is a rectangle. [4 MARKS]

Answer 1

Steps: 2 Marks
Each angle: 1 Mark

Since ABCD is a rectangle, therefore $AB\parallel CD$

$\angle EDC=\angle EIJ={50}^{\circ }$ (Corresponding Angles).

As ABCD is rectangle so $\angle ADC={90}^{\circ }$

$\Rightarrow \angle ADG+\angle GDC={90}^{\circ }$

So, $\angle ADG={90}^{\circ }-{50}^{\circ }={40}^{\circ }$.

$\angle EJI={180}^{\circ }-{105}^{\circ }={75}^{\circ }$ (Linear pairs of angles);

$\angle IEJ={180}^{\circ }-({75}^{\circ }+{50}^{\circ })$ (Angle sum property)

$\Rightarrow \angle IEJ={55}^{\circ }$.