MATHEMATICS PROBLEM SOLVER

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Friday, 14 February 2014

CIRCLES C4

In the figure below, find the value of <MNQ.

Solution

<MQN = <PMN.

∴ <MQN = 52⁰ [angles in alternate segments are equal.]

Consider ∆MQN

<QMN + <MNQ+ <MQN = 180⁰ [total degrees of a triangle]

(38⁰ + 52⁰) + <MNQ + 52= 180⁰

90⁰ + <MNQ + 52⁰ = 180⁰

142⁰ + <MNQ= 180⁰

<MNQ= 180⁰-142⁰

<MNQ= 38⁰

∴ <MNQ= 38⁰

MATHEMATICS PROBLEM SOLVER

Friday, 14 February 2014

CIRCLES C4

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