MATHEMATICS PROBLEM SOLVER

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Friday, 14 February 2014

CIRCLES C3

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

Solution

<ABD = <DAC

∴ <ABD = 51 [angles in alternate segments are equal.]

Consider ∆ABD

<ABD + <BAD+ <BDA = 180⁰ [total degrees of a triangle]

51 + (60 + 51) + <BDA= 180⁰

51 + 111 + <BDA= 180⁰

162 + <BDA= 180⁰

<BDA= 180⁰-162

<BDA= 18⁰

∴ <BDA= 18⁰

MATHEMATICS PROBLEM SOLVER

Friday, 14 February 2014

CIRCLES C3

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