Question

$ABCD$ is a square and $ABRS$ is a rhombus. if $\angle SAD={120}^{\circ }$,

find: (¡) $\angle ASD$ (¡¡) $\angle SRB$

Answer 1

Given:

$ABCD$ is a square.

$ABRS$ is a rhombus.

$\angle SAD={120}^{\circ }$

Lets form a diagram as per instructions,

Lets take $\Delta ASD$,

$AS=AD$ [Since, $ABCD$ is a square and $ABSR$ is a rhombus with side $AB$]

$\Rightarrow \Delta ASD$ is an isosceles triangle.

$\Rightarrow \angle ASD=\angle ADS$

$\Rightarrow \angle ASD+\angle ADS+\angle SAD={180}^{\circ }$

$\Rightarrow \angle ASD+\angle ASD+{120}^{\circ }={180}^{\circ }$

$\Rightarrow 2\angle ASD={180}^{\circ }-{120}^{\circ }$

$\Rightarrow 2\angle ASD={60}^{\circ }$

$\Rightarrow \angle ASD=\frac{{60}^{\circ }}{2}$

$\Rightarrow \angle ASD={30}^{\circ }$ ---(i)
It is given that,

$\angle SAD={120}^{\circ }$

$\Rightarrow \angle DAB+\angle SAB={120}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle SAB={120}^{\circ }$ [Since, $ABCD$ is a square.]

$\Rightarrow \angle SAB={120}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle SAB={30}^{\circ }$

In a Rhombus, opposite angles are equal.

$\Rightarrow \angle SAB=\angle SRB={30}^{\circ }$

$\therefore \angle SRB={30}^{\circ }$ ---(ii)