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In the Given Figure, Ab is Parallel to Dc ∠Bce = 80° and ∠Bac = 25°.Find: (I) ∠Cad (Ii) ∠Cbd (Iii) ∠Adc - ICSE Class 10 - Mathematics

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Question

In the given figure, AB is parallel to DC ∠BCE = 80° and ∠BAC = 25°.

Find:
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC

Solution

In the given figure,
ABCD is a cyclic quad in which AB ∥ DC
∴ ABCD is an isosceles trapezium
∴ AD = BC

Ext. ∠BCE = ∠BAD  [Exterior angle of a cyclic qud is equal to interior opposite angle]

 ∴ ∠BAD = 80°      [ ∵ ∠BCE = 80°]

But ∠BAC = 25°

∴ ∠CAD = ∠BAD - ∠BAC
           = 80° - 25°
            = 55°

(ii) ∠CBD = ∠CAD   [Angle of the same segment]
                = 55°

(iii) ∠ADC = ∠BCD   [Angles of the isosceles trapezium]

= 180° - ∠BCE
= 180° - 80°
= 100°

  Is there an error in this question or solution?
Solution In the Given Figure, Ab is Parallel to Dc ∠Bce = 80° and ∠Bac = 25°.Find: (I) ∠Cad (Ii) ∠Cbd (Iii) ∠Adc Concept: Cyclic Properties.
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