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AB is a diameter of the circle with cent...

AB is a diameter of the circle with centre `O` and chord `CD` is equal to radius `OC` (fig). `AC` and `BD` proudced meet at P. Prove that `angle CPD=60^(@)`.

A

`60^@`

B

`30^@`

C

`90^@`

D

`160^@`

Text Solution

Verified by Experts

The correct Answer is:
A
Since, chord CD is equal to radius
`therefore DC=OC=OD`
`rArrangle1 =60^@` (`because` each angle of an equilateral triangle is of `60^@`)
`thereforeangle2(1)/(2)angle1` (degree measure theorem)
`=(1)/(2)xx60^@=30^@` ........(1)
Now, since `angle3=90^@` (angle in a semicircle is `90^@`)
`rArrangle4=90^@` (linear pair axiom).....(2)
In `DeltaADP`,
`angle2+angle4+anle5=180^@` (angle sum property)
`rArr30^@+90^@+angle5=180^@`
`rArrangle5=180^@-120^@`
`60^@`
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