A Ferris wheel is an amusement ride consisting of a rotating upright wheel with multiple passenger carrying components
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, Sanjeev came
out from the crowd and was observing his friends
who were enjoying the ride. He was curious about
the different angles and measures that the wheel
will form. He forms the figure as given below.
M
60
d. 90°
Read the above passage carefully and give the
answer of the following questions.
Q1. In the given figure, find ZROQ.
a. 60 b. 100
C 120
Q2. Find ZORQ.
a. 15 b. 25 C. 30 d. 50°
Q3. Find ZRQP
a. 60° b. 75
C. 30° d. 90°
Q4. If the length of tangents PR = 4 cm and radius of
the circle is 3 cm, then length of OP is:
a 9 cm
b. 16 cm c. 5 cm
Q5. If the length of chord RQ = 4 cm, then PM is:
a. 2 cm b. 4 cm
c. 273 cm d. 73 cm
d. 25 cm
Answers
Step-by-step explanation:
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Answer :
The final answer is ∠RQO = 15° , ∠RQS = 180° – 150o = 30°
Explanation :
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°
⇒ ∠PRQ =
= 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° [Traingle rule 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° [Traingle rule Rule]
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°
To learn more about to find the angles of traingle refer :
https://brainly.in/question/35054263
https://brainly.in/question/19409067
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