The given figure shows a circle with center O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate :

(i) angle QTR

(ii) angle QRP

(iii) angle QRS

(iv) angle STR

(i) From figure,

⇒ ∠POQ + ∠QOR = 180° [Linear pairs]

⇒ 100° + ∠QOR = 180°

⇒ ∠QOR = 180° - 100°

⇒ ∠QOR = 80°.

We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc RQ subtends ∠QOR at the center and ∠QTR at the remaining part of the circle.

⇒ ∠QOR = 2∠QTR

⇒ ∠QTR = $\dfrac{1}{2}$∠QOR = $\dfrac{1}{2} \times 80° = 40°.$

Hence, ∠QTR = 40°.

(ii) We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc QP subtends ∠QOP at the center and ∠QRP at the remaining part of the circle.

⇒ ∠QOP = 2∠QRP

⇒ ∠QRP = $\dfrac{1}{2}$∠QOP = $\dfrac{1}{2} \times 100° = 50°.$

Hence, ∠QRP = 50°.

(iii) Given,

RS || QT

⇒ ∠SRT = ∠QTR = 40° (Alternate angles are equal)

From figure,

∠QRS = ∠QRP + ∠PRT + ∠SRT = 50° + 20° + 40° = 110°.

Hence, ∠QRS = 110°.

(iv) Since, RSTQ is a cyclic quadrilateral and sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠QRS + ∠QTS = 180°

⇒ ∠QRS + ∠QTR + ∠STR = 180°

⇒ 110° + 40° + ∠STR = 180°

⇒ ∠STR = 180° - 150°

⇒ ∠STR = 30°.

Hence, ∠STR = 30°.