Question

In the above figure (not to scale), O is the centre
of the circle. BC and CD are equal chords. If angle OBC = 70°, then find angle BAD.​

Answer 1

➩Given

❁ O is the centre of the circle

❁ BC=CD (Equal chords)

❁∠OBC=70°

➩To Find

∠BAD=?

➩Solution

In △BOC

OB=OC (Radius)

∠OBC=∠OCB (Angles opposite to equal sides are always equal)

so ∠OCB=70° (as ∠OBC=70°)

Also, In △OBC

∠OBC+∠OCB+∠BOC=180° (Sum of all the three angles of a triangle is 180°)

70°+70°+∠BOC=180°

140°+∠BOC=180°

∠BOC=180°-140°

∠BOC=40°

Now, in △OBC and △ODC

OB=OD (Radius)

OC=OC (Common)

BC=CD (Given)

△OBC≅△ODC (by SSS criterion)

So,

∠BOC=∠DOC

∠OBC=∠ODC

∠OCB=∠OCD (by CPCT)

Now, ∠BOC+∠DOC=∠BOD

40°+40°=80°

∠BOD=80°

→We know angle subtended by an arc at the centre is double the angle subtended by the same arc at the remaining part of the circle

∠BOD=2∠BAD

$\frac{1}{2}\times∠BOD =∠BAD$

$\frac{1}{2}\times80 = 40°$

Therefore ,∠BAD=40°

Answer 2

answer

Step-by-step explanation:

OB=OC(Radius of same circle(

angle OCB=70

angle BOC=40(Angle sum property)

angle BOD=80

BAD=1/2*80

=40