Question

In the figure given below,AB is the diameter of a circle with centre O.Find Angle BDC.​

Answer 1

Answer:

∠BDC=30°

Step-by-step explanation:

∠ADC=1/2*∠AOC

=1/2*120=60

Now in ΔADC

∠ADC=90° (ANGLE IN SEMICIRCLE)

and ∠BDC=∠ADB-∠ADC

=90°-60°

∠BDC=30°

Answer 2

Given : ∠AOC = 120°

To find : ∠BDC

Solution:

∠AOC = 120°

now ,

∠AOC +∠BOC = 180° (linear pair)

120° +∠BOC = 180°

∠BOC = 180°- 120°

∠BOC = 60°

now , we know that angle at the center is twice the angle at the circumference subtended by the same arc

therefore ,

$\angle BDC = \frac{1}{2}\angle BOC \\\\ \angle BDC = \frac{1}{2}\times 60^\circ \\\\\angle BDC = 30^\circ$

hence The measure of ∠BDC is 60°

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