Question

AB||CD
angle APQ = 60° and angle PRD = 125°
find the value of X
pls answer ​

Answer 1

Answer:

Ans:60°

Apq:60° alternative angles

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Answer 2

$x \: = \: 60^{\circ}$

GIVEN :

[•

\sf AB \parallel CD[/tex]

$• </p><p>\sf \angle APQ = 60^{\circ}$

$• </p><p>\sf \angle PRD = 137^{\circ}$

TO FIND :

The values of x

SOLUTION :

From the given figure,

R is a point angle. So,

$\angle R = 180^{\circ}$

$→ </p><p>\angle QRP \: + \: \angle PRD \: = \: 180^{\circ}$

$→ </p><p>\angle QRP \: = \: 180^{\circ} \: - \: 137^{\circ}$

$→ </p><p>\angle QRP \: = \: 43^{\circ}$

$In </p><p>\triangle PQR,$

$→ </p><p>x \: + \: y \: + \: 43^{\circ} \: = \: 180^{\circ}$

$→ </p><p>x \: + \: y \: = \: 180^{\circ} \: - \: 43^{\circ}$

$→ </p><p>x \: + \: y \: = \: 137^{\circ}$

$\therefore AB \parallel CD, \angle APQ \: and \: \angle PQR \: are \: alternate \: angles.$

$Now, </p><p>\angle APQ \: = \: \angle PQR$

$\therefore \sf {\boxed {x \: = \: 60^{\circ}}}$