Question

In given figure, find the values of $x$ and $y$, if $QS=RQ$.

• A. $x=36{}^{\circ },y=32{}^{\circ }$
• B. $x=45{}^{\circ },y=32{}^{\circ }$
• C. $x=32{}^{\circ },y=45{}^{\circ }$
• D. $x=45{}^{\circ },y=26{}^{\circ }$

Answer 1

The correct option is D $x=45{}^{\circ },y=26{}^{\circ }$

Given,

$\angle XPS={148}^{\circ }$.

Since $XPQ$ is a straight line,

$⟹\angle XPS+\angle QPS={180}^{\circ }$ ....(LInear pairs)

$\therefore \angle QPS={180}^{\circ }-{148}^{\circ }={32}^{\circ }$.

In $△SPQ$

$\angle S+\angle P+\angle Q={180}^{\circ }$ ......(Angle sum property)

$⟹$ ${103}^{\circ }+{32}^{\circ }+x^{\circ }={180}^{\circ }$

$⟹$$x^{\circ }={45}^{\circ }$.

Given,

$\angle PSQ={103}^{\circ }$.

Since $PSR$ is a straight line,

$⟹\angle PSQ+\angle QSR={180}^{\circ }$

$\therefore \angle QSR={180}^{\circ }-{103}^{\circ }={77}^{\circ }$.

In $△SQR$,

given $QS=RQ$.

$⟹△SQR$ is isosceles.

By Isosceles Triangle property, the angles opposite to equal sides are also equal.

$⟹$$\angle SRQ=\angle QSR={77}^{\circ }$

$⟹$$\angle S+\angle Q+\angle R={180}^{\circ }$

$⟹$${77}^{\circ }+y^{\circ }+{77}^{\circ }={180}^{\circ }$

$⟹$${154}^{\circ }+y^o={180}^{\circ }$

$⟹$$y^o={180}^{\circ }-{154}^o$

$⟹$$y^{\circ }={26}^{\circ }$.