Question

In the given figure. O in the centre of the circle and AB is the diameter. If angleODA = 90 and 0B - BC find 21 D B ) 30" C) 60° d) 70​

Answer 1

Answer:

For chord CD,

∠CBD=∠CAD ...Angles in same segment

∠CAD=70°

∠BAD=∠BAC+∠CAD=30°+70°=100°

∠BCD+∠BAD=180° ...Opposite angles of a cyclic quadrilateral

⇒∠BCD+100°=180°

⇒∠BCD=80°

In △ABC

AB=BC (given)

∠BCA=∠CAB ...Angles opposite to equal sides of a triangle

∠BCA=30°

Also, ∠BCD=80°

∠BCA+∠ACD=80°

⇒30°+∠ACD=80°

∠ACD=50°

∠ECD=50°

solution

Answer 2

In the given figure, $O$ is the centre of the circle and $\mathrm{AB}$ is the diameter. If $\angle O D A=90^{\circ}$ and $\mathrm{OB}= B C$, find $\angle 1$.

$21 D B ) 30" C) 60\ {circ} d) 70​$

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