MATHEMATICS PROBLEM SOLVER

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Monday, 24 February 2014

CIRCLES C5

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

Solution

<HJL = <LHK

∴ <HJL = 54⁰ [angles in alternate segments are equal.]

Consider ∆HJL

<JHL + <HLJ+ <HJL = 180⁰ [total degrees of a triangle]

(44⁰ + 54⁰) + <HLJ + 54= 180⁰

98⁰ + <HLJ + 54⁰ = 180⁰

152⁰ + <HLJ= 180⁰

<HLJ= 180⁰- 152⁰

<HLJ= 28⁰

∴ <HLJ= 28⁰

MATHEMATICS PROBLEM SOLVER

Monday, 24 February 2014

CIRCLES C5

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