Question

In the given figure if AB parallel CD, angle APQ=60°and angle PRD=37°, then find the value of x and y

Answer 1

The value of $x$ and $y$ is $60\ degree$ and $77\ degree$

Step-by-step explanation:

Given,

In figure, $AB \parallel CD$,$\angle APQ=60\ degree$ and $\angle PRD=137\ degree$

From figure,

$R$ is a point angle.

So, $\angle R=180\ degree$

⇒$\angle QRP+\angle PRD=180$

⇒$\angle QRP=180-137$

⇒$\angle QRP=43$

In $\triangle PQR$,

$x+y+43=180$

⇒$x+y=180-43$

⇒$x+y=137$__1

∵ $AB \parallel CD$, $\angle APQ$ and $\angle PQR$ alternate angles.

So, $\angle APQ=\angle PQR$

⇒$x=60\ degree$

Plug the value in equation-1,

⇒$60+y=137$

⇒$y=137-60$

∴ $y=77\ degree$

So, The value of $x$ and $y$ is $60\ degree$ and $77\ degree$

Answer 2

Given :  AB is parallel to CD, angle APQ=60 ° and angle PRD =137°

To Find :  value of x and y

Solution:

Given AB || CD    and PQ is transversal

Hence  x    = 60°    ( alternate angle )

137°  = x + y       as Exterior angle of triangle = sum of two opposite interior angles

=>  137°  = 60°  + y  

=> y = 77°

x =  60°   and y = 77°

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