TS ICET 23 May 2018 Shift 1 Paper

Question : 138 of 200

Marks: +1, -0

A. B, C. D are four points on a circle AC and BD intersect at E such that ∠BEC= 130° and ∠ECD = 20°. Then ∠BAC'=

Solution:

Given ∠BEC= 130° and ∠ECD = 20°.

∠\CED=180°-∠\BEC=180-130=50°

∆CED,

∠\EDC=180-∠\CED-∠\ECD

=180-50-20=110°

Required angle.

∠\BAC = ∠\BDC = ∠\EDC = 110°

(Since chord BC subtends equal angles at points A and D)

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