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In the Figure, Given Below, Ad = Bc, ∠Bac = 30° and ∠Cbd = 70°. Find: (I) ∠Bcd (Ii) ∠Bca (Iii) ∠Abc (Iv) ∠Adb - Mathematics

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Question

In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°.

Find: 

  1. ∠BCD
  2. ∠BCA
  3. ∠ABC
  4. ∠ADB

Sum

Solution

In the figure, ABCD is a cyclic quadrilateral

AC and BD are its diagonals

∠BAC = 30° and ∠CBD = 70° 

Now we have to find the measure of

∠BCD, ∠BCA, ∠ABC and ∠ADB

We have ∠CAD = ∠CBD = 70°  ...[Angles in the same segment]

Similarly, ∠BAD = ∠BDC = 30°

∴ ∠BAD = ∠BAC + ∠CAD

= 30° + 70°

= 100° 

i. Now ∠BCD + ∠BAD = 180°   ...[Opposite angles of cyclic quadrilateral]

`=>` ∠BCD + ∠BAD = 180°

`=>` ∠BCD + 100° = 180°

`=>` ∠BCD = 180° – 100°

`=>` ∠BCD = 80° 

ii. Since AD = BC,

ABCD is an isosceles trapezium and AB || DC

∠BAC = ∠DCA    ...[Alternate angles]

`=>` ∠DCA = 30°   

∴ ∠ABD = ∠DAC = 30°  ...[Angles in the same segment]

∴ ∠BCA = ∠BCD – ∠DAC

= 80° – 30°

= 50° 

iii. ∠ABC = ∠ABD + ∠CBD

= 30° + 70° 

= 100°

iv. ∠ADB = ∠BCA = 50°   ...[Angles in the same segment]

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Chapter 17: Circles - Exercise 17 (C) [Page 267]

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Selina Mathematics [English] Class 10 ICSE
Chapter 17 Circles
Exercise 17 (C) | Q 23 | Page 267

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