help me with this.........
Answers
Step-by-step explanation:
Hey Mate...!!!✌
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\bold{\Huge\red{\boxed{{{Answer}}}}}
Answer
➤ \: < /p > < p > < /p > < p > \angle{BPR}=70°➤</p><p></p><p>∠BPR=70°
➤ \: < /p > < p > < /p > < p > \angle{QPR}=80°➤</p><p></p><p>∠QPR=80°
\bold{\Huge\purple{\boxed{{{Given}}}}}
Given
➤ \: < /p > < p > < /p > < p > \angle{APQ}=50°➤</p><p></p><p>∠APQ=50°
➤ \: < /p > < p > < /p > < p > \angle{PRD}=130°➤</p><p></p><p>∠PRD=130°
➤ < /p > < p > < /p > < p > \: AB \: || \: CD➤</p><p></p><p>AB∣∣CD
\bold{\Huge\green{\boxed{{{To\: Find}}}}}
ToFind
➤ \: < /p > < p > < /p > < p > \angle{BPR}➤</p><p></p><p>∠BPR
➤ \: < /p > < p > < /p > < p > \angle{QPR}➤</p><p></p><p>∠QPR
\bold{\Huge\blue{\boxed{{{Explanation}}}}}
Explanation
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➠ As line AB || CD,
{\angle{APQ}=\angle{PQR}}\: {{ (Alternate Angles)}}∠APQ=∠PQR(AlternateAngles)
= > \angle{PQR} = 50°=>∠PQR=50°
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➠ On line QD,
= > \angle{PRD} + \angle{PRQ} = 180°=>∠PRD+∠PRQ=180°
= > 130° + \angle{PRQ} = 180°=>130°+∠PRQ=180°
= > \angle{PRQ} = 180° - 130°=>∠PRQ=180°−130°
= > \angle{PRQ} = 50°=>∠PRQ=50°
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➠ In ΔPQR,
\small{{= > \angle{PRQ} + \angle{PRD} + \angle{QPR} = 180° (Angle \: Sum \: Property \: of \: Δ)}}=>∠PRQ+∠PRD+∠QPR=180°(AngleSumPropertyofΔ)
= > 50° + 50° + \angle{QPR} = 180°=>50°+50°+∠QPR=180°
= > 100° + \angle{QPR} = 180°=>100°+∠QPR=180°
= > \angle{QPR} = 180° - 100°=>∠QPR=180°−100°
\boxed{\bold{= > \angle{QPR} = 80°}}
=>∠QPR=80°
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➠ On line AB,
\small{= > \angle{QPR} + \angle{APQ} + \angle{BPR} = 180° (Linear \: Pair)}=>∠QPR+∠APQ+∠BPR=180°(LinearPair)
= > 80° + 50° + \angle{BPR} = 180°=>80°+50°+∠BPR=180°
= > 130° + \angle{BPR} = 180°=>130°+∠BPR=180°
= > \angle{BPR} = 180° - 130°=>∠BPR=180°−130°
\boxed{\bold{= > \angle{BPR} = 50°}}
=>∠BPR=50°
Hope It Helps You....❤