#### Question

In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.

Calculate:

(i) ∠RNM,

(ii) ∠NRM

#### Solution

(i) Join RN and MS.

∴ ∠RMS = 90°

(Angle in a semicircle is a right angle)

∴ ∠RSM = 90° - 29° = 61°

(By angle sum property of triangle RMS)

∴ ∠RNM =180° ∠RSM =180° - 61° = 119°

(pair of opposite angles in a cyclic quadrilateral are supplementary)

(ii) Also, RS || NM

∴ ∠NMR = ∠MRS = 29° (Alternate angles)

∴ ∠NMS = 90° + 29° = 119°

Also, ∠NRS + ∠MS = 180°

(pair of opposite angles in a cyclic quadrilateral are supplementary)

⇒ ∠NMR + 29° +119° = 180°

⇒ ∠NRM = 180° -148°

∴ ∠NRM = 32°

Is there an error in this question or solution?

Solution In the Given Figure, Rs is a Diameter of the Circle. Nm is Parallel to Rs and ∠Mrs = 29°. Calculate: (I) ∠Rnm, (Ii) ∠Nrm Concept: Arc and Chord Properties - Angle in a Semi-circle is a Right Angle.