Q) 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.

1. In the given figure find ∠ ROQ
a) 60            b) 100                 c) 150                 d) 90

2. Find ∠ RQP
a) 75            b) 60                  c) 30                   d) 90

3. Find ∠ RSQ
a) 60           b) 75                  c) 100                  d) 30

4. Find ∠ ORP
a) 90           b) 70                   c) 100                  d) 60 

Ans: 

1. ∠ ROQ: 

Step 1: In Quadrilateral PQOR, sum of all angles is 360

∴ ∠ PRO + ∠ ROQ + ∠ OQP + ∠ QPR = 360 ……….. (i)

Step 2:  Since radii OQ and OR are perpendicular to the tangents,

∴ ∠ OQP = ∠ PRO = 90

∠ QPR = 30 (given)

Step 3: By substituting these values in equation (i), we get:

90 + ∠ ROQ + ∠  90 + 30 = 360

∴ ∠  ROQ = 360 – 210

∴ ∠  ROQ = 150

Therefore, option (c) is correct.

2. ∠ RQP: 

Step 4: ∵ Sum of all angles of a triangle is 180

∴  ∠ OQR + ∠ ORQ + ∠ ROQ = 180 ……….. (ii)

Step 5: In Δ QOR, OQ = OR   (radii of a circle)

∴ ∠ OQR = ∠ ORQ                  (angle opposite to equal sides)

Also ∠ ROQ = 150                   (from step 3 above)

Step 6: By substituting these 2 values in equation (ii), we get:

∴  ∠ OQR + ∠ OQR + ∠ ROQ = 180         

∴  2 ∠ OQR + 150 = 180                           

∴  2 ∠ OQR =  180  –  150 = 30

∴  ∠ OQR = = 15

Step 7:

∵ ∠ OQP = ∠ OQR  + ∠ RQP 

Step 8: Here, ∠ OQP = 90            (radius OQ is to tangent PQ)

and ∠ OQR = 15                            (calculated in Step 6 above)

Step 9: By substituting these values in equation (i), we get:

∵ ∠ OQP = ∠ OQR  + ∠ RQP 

∴  90 = 15 + ∠ OQR

∴ ∠ OQR = 90 – 15

∴ ∠ OQR = 75

Therefore, option (a) is correct.

3. ∠ RSQ:

Step 10: By inscribed angle theorem, we know that “the angle subtended by an arc at the center of a circle is twice the angle subtended by the arc at any point on the circumference.”

∴ ∠ ROQ = 2 ∠ RSQ

∴ ∠ RSQ = ∠ ROQ

Step 11: ∠ ROQ = 150              (from step 3 above)

Step 12: By substituting these values in equation (i), we get:

∵ ∠ RSQ = ∠ ROQ

∴ ∠ RSQ = (150)

∴ ∠ RSQ = 75

Therefore, option (a) is correct.

4. ∠ ORP:

Step 13:  Since radius OR is perpendicular to the tangent PR,

∴ ∠ ORP = 90

Therefore, option (a) is correct.

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