Question

In the figure, given below, AD = BC, angle BAC
= 30° and angle CBD = 70°.
Find :
(i) angle BCD
(ii) angle BCA
(iii) angle ABC
(iv) angle ADB
Plz help me sm1
I will mark u BRAINLIEST
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Answer 1

Answer:

BCD IS 90, BCA IS 60, ABC IS 90, ADB IS 70

Answer 2

Step-by-step explanation:

Given that:

AD = BC, angle BAC

= 30° and angle CBD = 70°.

To find:

Find :

(i) angle BCD

(ii) angle BCA

(iii) angle ABC

(iv) angle ADB

Solution:

1) ABCD is a cyclic quadrilateral

(Property of cyclic quadrilateral: Sum of opposite angles are 180°)

2) AD=BC

(Equal chords subtends equal angles on the same segment of circle)

(iv) angle ADB

$\angle \: BAC = \angle \: BDC \: \: \: (from \: 2)\\ \\ \angle \: BAC = 30° \\ so \\ \angle \: BDC = 30° \\ \\ \angle \: ADC = \angle \: BAC(from \: 2)\\ \\ \angle \: ADC = 30° = \angle \: ADB$

(iii) angle ABC

Now,

$\angle \: ABC = \angle \: ABD + \angle \: DBC \\ \\ \angle \: ABC = 30° + 70° \\ \\ \angle \: ABC = 100°$

(iv) angle ADB

So,

from property 1

$\angle \: ADC + \angle \: ABC = 180° \\ \\ \angle \: ADC = 180° - 100° \\ \\ \angle \: ADC = 80°$

So,

$\angle \: ADB= 80° - 30° \\ \\ \angle \: ADB = 50°$

(ii) angle BCA

Let the two dashed lines meets at O

So,

$\angle \: AOB = \angle \: COD \: \: V.O.A) \\ \\ in \: triangle,\\\\\: AOB \angle \: AOB = 120° =\angle \: COD \\ \\ so \\ \\ \angle \: BOC = 60° \\ \\ in \: triangle \: BOC \\ \\ \angle \: BCA = 180° - \angle \: COD - \angle \: OBC \\ \\ \angle \: BCA = 180° - 60°- 70° \\ \\ \angle \: BCA =50°$

Hope it helps you.