In the figure, AC is perpendicular to CE and ∠A: ∠B: ∠C = 3:2:1, the measure of ∠ECD is

80^o

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40^o

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90^o

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60^o

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Solution

The correct option is D
60^o

Let the angles be 3x, 2x, x

By angle sum property of a triangle you get

x= 30^o

Hence ÐA = 90^o, ÐB = 60^o and ÐACB =30^o.

Now, ÐACB + ÐACE + ÐECD = 180^o (Angles on a straight line)

ÐECD = 60^o

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∠ B = 60^{\circ}

$I n a Δ A B C, it is given that ∠ A : ∠ B : ∠ C = 3 : 2 : 1 a n d ∠ A C D = 90^{\circ} . If BC is produced to E then ∠ E C D = ?$

$(a) 60^{\circ} (b) 50^{\circ} (c) 40^{\circ} (d) 25^{\circ}$

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