Advertisements
Advertisements
प्रश्न
The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is
विकल्प
60°
75°
120°
150°
Advertisements
उत्तर
150°
We are given that the chord is equal to its radius.
We have to find the angle subtended by this chord at the minor arc.
We have the corresponding figure as follows:

We are given that
AO = OB = AB
So ,
\[\bigtriangleup\] AOB is an equilateral triangle.
Therefore, we have
∠AOB = 60°
Since, the angle subtended by any chord at the centre is twice of the angle subtended at any point on the circle.
So `angleAQB =(angleAOB)/2`
`= 60/2 = 30°`
Take a point P on the minor arc.
Since `square APBQ` is a cyclic quadrilateral
So, opposite angles are supplementary. That is
`angle APB + angleAQB = 180°`
`angle APB + 30° = 180°`
`angleAPB = 180° - 30°`
`= 150°`
APPEARS IN
संबंधित प्रश्न
Fill in the blank:
A circle divides the plane, on which it lies, in ............ parts.
Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.
In the below fig. O is the centre of the circle. Find ∠BAC.

If O is the centre of the circle, find the value of x in the following figure

In the given figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = BC.

In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.

If ABC is an arc of a circle and ∠ABC = 135°, then the ratio of arc \[\stackrel\frown{ABC}\] to the circumference is ______.
In the following figure, ∠ACB = 40º. Find ∠OAB.

A circle has radius `sqrt(2)` cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45°.
