In figure below,AB"CD,angleAPQ=50,anglePRD=127,find the value of x and respectively are
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Given:
AB║CD
∠APQ = 50°
∠PRD = 127°
To find:
∠x, and ∠y.
Solution:
ATQ, AB║CD. Considering PQ as the transversal through AB & CD, we can say that:
⇒ ∠APQ = ∠x
(Alternate interior angles are equal)
⇒ 50° = ∠x
∴ ∠x = 50°
AB║CD & when we consider PR as the transversal, we can say that:
⇒ ∠BPR + ∠DRP = 180° (Co-interior angles)
⇒ ∠BPR + 127° = 180°
⇒ ∠BPR = 180° - 127°
⇒ ∠BPR = 53°
On line APB,
⇒ ∠APQ + ∠QPR + ∠BPR = 180° (Straight angle)
⇒ 50° + ∠y + 53° = 180°
⇒ ∠y + 103° = 180°
⇒ ∠y = 180° - 103°
⇒ ∠y = 77°
Final answers:
∠x = 50°
∠y = 77°
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