Question

Answer the question in the attachment

Answer 1

Step-by-step explanation:

In ∆OAB,

E is the mid-point of OB

And, D is the mid-point of AB

Therefore, By mid-point theorem, we get :-

ED||OA and ED = 1/2 OA

Also, F is the mid-point of OA

And, D is the mid-point of AB

Therefore, By mid-point theorem, we get :-

FD||OB and FD = 1/2 OB

OA = OB (radii of the circle)

Dividing both sides by 1/2, we get :-

1/2 OA = 1/2 OB

Therefore, ED = FD (Since, ED = 1/2 OA and FD = 1/2 OB)

So, Angle EFD = Angle DEF = 30° (Since, angles opposite to equal sides are equal).........(i)

Also, In ∆DEF,

Angle DEF = 30° (given)

Angle EFD = 30° [ using (i) ]

Angle FDE = ?

We know that,

Angle DEF + Angle EFD + Angle FDE = 180° (A.S.P of a ∆)

=> 30° + 30° + Angle FDE = 180°

=> 60° + Angle FDE = 180°

Therefore, Angle FDE = 180° - 60°

= 120°

Also, In ∆OEF,

OE = 1/2 OB (Since, E is the mid-point of OB)

OF = 1/2 OA (Since, F is the mid-point of OA)

Now, In ∆OEF and ∆DEF,

EF = EF (common)

OE = FD (Since, OE = FD = 1/2 OB)

OF = ED (Since, OF = ED = 1/2 OA)

Therefore, ∆OEF ≅ ∆DEF ( By S.A.S congruency rule)

Also, Angle FOE = Angle FDE = 120° (By C.P.C.T)

Therefore, Angle ACB = 1/2 of Angle FOE ( Since, angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of the circle)

So, Angle ACB = 1/2 × Angle FOE

= 1/2 × 120°

= 60°

Hope it helps :)