In the diagram, $O$ is the centre of the circle and $D, E$ and $F$ are mid points of $A B$ , $B O$ and $O A$ , respectively. If $∠ D E F = 30^{0}$ , the $∠ A C B$ is

30^{0}

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60^{0}

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90^{0}

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120^{0}

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Solution

The correct option is A $60^{0}$
ADEF is a parallelogram. Opposite angles in parallelogram are equal so $∠ F A D = 30^{0} .$
Since OAB is an isosceles triangle, $∠ A O B = 120^{0}$
Thus $∠ A C B = 0.5 \times ∠ A O B = 60^{0} .$

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Similar questions

Assertion (A) in the given figure, O is the centre of the circle with D, E and F as mid-points of AB, BO and OA respectively. If $∠ D E F$ is $30^{\circ}$ , then $∠ A C B$ is $60^{\circ}$

Reason (R) Angle subtended by an arc at the centre is twice the angle subtended by it on the remaining part of the circle.
which of the following is true?

Q. If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that

\vec{O A} + \vec{O B} + \vec{O C} = \vec{O D} + \vec{O E} + \vec{O F}

Q. In the given figure of

△ A B C

, let

∠ A = 50^{o}, B = 60^{o},

and

C = 70^{o}

. If mid-point of side

A B, B C

and

C A

are

D, E

and

F

respectively then find

∠ D E F

Equal chords AB and CD of a circle with centre O, cut at right angles at E. If M and N are the mid - points of AB and CD respectively, then OMEN is a __.

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In the diagram, O is the centre of the circle and D,E and F are mid points of AB, BO and OA, respectively. If ∠DEF=300, the ∠ACB is

The correct option is A 600ADEF is a parallelogram. Opposite angles in parallelogram are equal so ∠FAD=300.Since OAB is an isosceles triangle, ∠AOB=1200Thus ∠ACB=0.5×∠AOB=600.

In the diagram, $O$ is the centre of the circle and $D, E$ and $F$ are mid points of $A B$ , $B O$ and $O A$ , respectively. If $∠ D E F = 30^{0}$ , the $∠ A C B$ is

The correct option is A $60^{0}$
ADEF is a parallelogram. Opposite angles in parallelogram are equal so $∠ F A D = 30^{0} .$
Since OAB is an isosceles triangle, $∠ A O B = 120^{0}$
Thus $∠ A C B = 0.5 \times ∠ A O B = 60^{0} .$