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A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.
After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.
Q. Find ∠ORP
  • a)
    90
  • b)
    70
  • c)
    100
  • d)
    60
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement...
PR = PO [∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR [∵ Angles opposite equal sides are equal
In ∆PQR,
⇒ ∠PRQ + ∠RPQ + ∠POR = 180° [∆ Rule]
⇒ 30° + 2∠PQR = 180°
= 75° ⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR [Alternate interior angle]
∴ ∠SRO = 75° [Tangent is I to the radius through the point of contact]
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° [∆ Rule]
∴ ∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150°
⇒ ∠QSR = 12∠QO
⇒ ∠QSR = 150/2 = 750 [Used ∠SRQ = 75° as solved above]
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° [∆ Rule]
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°
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Most Upvoted Answer
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement...
PR = PO [∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR [∵ Angles opposite equal sides are equal
In ∆PQR,
⇒ ∠PRQ + ∠RPQ + ∠POR = 180° [∆ Rule]
⇒ 30° + 2∠PQR = 180°
= 75° ⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR [Alternate interior angle]
∴ ∠SRO = 75° [Tangent is I to the radius through the point of contact]
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° [∆ Rule]
∴ ∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150°
⇒ ∠QSR = 12∠QO
⇒ ∠QSR = 150/2 = 750 [Used ∠SRQ = 75° as solved above]
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° [∆ Rule]
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°
Free Test
Community Answer
A Ferris wheel (or a big wheel in the United Kingdom) is an amusement...
PR = PO [∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR [∵ Angles opposite equal sides are equal
In ∆PQR,
⇒ ∠PRQ + ∠RPQ + ∠POR = 180° [∆ Rule]
⇒ 30° + 2∠PQR = 180°
= 75° ⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR [Alternate interior angle]
∴ ∠SRO = 75° [Tangent is I to the radius through the point of contact]
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° [∆ Rule]
∴ ∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150°
⇒ ∠QSR = 12∠QO
⇒ ∠QSR = 150/2 = 750 [Used ∠SRQ = 75° as solved above]
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° [∆ Rule]
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°
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A Ferris wheel (or a big wheel in the United Kingdom) is an amusement ride consisting of a rotating upright wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas, or pods) attached to the rim in such a way that as the wheel turns, they are kept upright, usually by gravity.After taking a ride in Ferris wheel, Aarti came out from the crowd and was observing her friends who were enjoying the ride . She was curious about the different angles and measures that the wheel will form. She forms the figure as given below.Q. Find ∠ORPa)90b)70c)100d)60Correct answer is option 'A'. Can you explain this answer?
Question Description
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