In the figure , AB and CD are parallel lines and transversal EF intersect them at P and Q . If < APR =250 , < RQC = 300 and < CQF = 650 , then find the measure of < RPQ and< PRQ.
Answers
Given :-
- AB and CD are parallel lines and transversal EF intersect them at P and Q .
- ∠APR =25°
- ∠RQC = 30°
- ∠CQF = 65°
To find :-
- ∠RPQ
- ∠PRQ
Construction :-
- Draw a line ST parallel to AB .
Solution :-
→ AB || ST
→ AB || CD
So,
→ CD || ST { Two lines parallel to the same line are parallel to each other. }
Now, AB || ST and EF is the transversal, So,
→ ∠APR = ∠PRT { Alternate interior angles .}
→ ∠PRT = 25° ----------- Eqn.(1)
Similarly, CD || ST and EF is the transversal,
→ ∠QRT = ∠RQC
→ ∠QRT = 30° ----------- Eqn.(2)
Adding Eqn.(1) and Eqn.(2), we get,
→ ∠PRT + ∠QRT = 25° + 30°
→ ∠PRQ = 55° (Ans.)
Now, FPE is a straight line, So,
→ ∠CQF + ∠CQR + ∠RQP = 180° .{ Angles forming a linear pair are supplementary.}
→ 65° + 30° + ∠RQP = 180°
→ ∠RQP = 180° - 95°
→ ∠RQP = 85°
Now, in ΔPRQ, we have :-
→ ∠RPQ + 55° + 85° = 180° {Angle sum Property.}
→ ∠RPQ + 140° = 180°
→ ∠RPQ = 180° - 140°
→ ∠RPQ = 40° (Ans.)
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