Question

(a) In the figure (i) given below, PQ is a diameter. Chord SR is parallel to PQ. Given
ZPQR = 58°, calculate
(ii) ZSTP
i
(1) ZRPQ 32​

Answer 1

Answer:

i) 90°

ii) 148°

Step-by-step explanation:

(i) ∠PRQ = 90˚ (∠ in a semi-circle)

In △PQR,

∠RPQ + ∠PQR + PRQ = 180˚ (∵ The sum of the three ∠s of a △ 180˚)

⇒ ∠RPQ + 58˚ + 90˚ = 180˚

⇒ ∠RPQ + 148˚ = 180˚

⇒ ∠RPQ = 180˚ – 148˚

⇒ ∠RPQ = 32˚

(ii) ∵ PQ ∥ SR and RP intersect them

∠PRS = ∠RPQ (Alternate angles)

∴ ∠PRS = 32˚

∵ PTSR is a cyclic quadrilateral.

∴ ∠PTS + ∠PRS = 180˚ (∵ Opposite ∠s of a cyclic quadrilateral are supplementary )

⇒ ∠PTS + 32˚ = 180˚

⇒ ∠PTS = 180˚ – 32˚ = 148˚

⇒ ∠STP = 148˚...

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Answer 2

Step-by-step explanation:

angle stp + angle srp = 180. (co interior property of cyclic quadrilateral)

<stp = 180-32 = 148

Answer