In the given figure, what is the angle subtended by chord AB at the point C on the circle?

60^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

65^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

55^{\circ}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

70^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $55^{\circ}$
Note that $∠ A C B$ is the angle subtended by the chord AB at the point C on the circle.
Using angle sum property in $△ A B C,$ we have
$∠ A B C + ∠ A C B + ∠ B A C = 180^{\circ} .$
$⟹ 60^{\circ} + ∠ A C B + 65^{\circ} = 180^{\circ}$
$⟹ ∠ A C B = 180^{\circ} - 125^{\circ} = 55^{\circ}$

Suggest Corrections

Similar questions

Q. In the given figure, what is the angle subtended by chord AB at the point C on the circle?

O is the centre of the circle. AB is a minor arc of the circle. The angle subtended by AB at centre $∠$ AOB $= 110^{\circ}$ , then angle subtended by the arc at any point on the circle $∠$ APB is

[P is any point on the circle]

In the figure, PC is the tangent to the circle. A and B are 2 points on the circle. If $∠$ BPC $= 60^{0}$ and $∠$ APB $= 55^{0}$ , then find $∠$ ABP .

In the figure given, the radius of the circle is 15 cm. The angle subtended by the chord AB at the centre O is

60^{\circ}

. What is the area of the minor segment?

In the given figure, $A B$ and $C D$ are two chords of a circle, intersecting each other at a point $E .$ Prove that $∠ A E C = \frac{1}{2}$ (angle subtended by arc $C X A$ at the centre $+$ angle subtended by arc $D Y B$ at the centre).

Join BYJU'S Learning Program

In the given figure, what is the angle subtended by chord AB at the point C on the circle?

The correct option is C 55∘Note that ∠ACB is the angle subtended by the chord AB at the point C on the circle. Using angle sum property in △ABC, we have ∠ABC+∠ACB+∠BAC=180∘. ⟹60∘+∠ACB+65∘=180∘ ⟹∠ACB=180∘−125∘=55∘