The tangent at a point C of a circle and a diameter AB when extended intersect at P. If `anglePCA=110^(@) "find" angleCBA`.

Options

A. `20^@`
B. `70^@`
C. `60^@`
D. `80^@`

Answer

B

Explanation

Here, AB is a diameter of the circle from point C and a tangent is drawn which meets at a point P.

Join OC. Here, OC is radius.
Since, tangent at any point of a circle is perpendicular to the radius through point of contact circle.
`OCbotPC`
Now, `anglePCA=110^(@)` [given]
`impliesanglePCO+angleOCA=110^(@)`
`implies90^(@)+angleOCA=110^(@)`
`impliesangleOCA=20^(@)`
`:.OC=OA="Radius of circle"`
` impliesangleOCA=angleOAC=20^(@)`
[since, two sides are equal, then their opposite angles are equal]
Sicne, PC is a tangent, so `angleBCP=angleCAB=20^(@)`
[angles in a alternate segment are equal]
In `DeltaPBC, angleP+angleC+angleA=180^(@)`
`angleP=180^(@)-(angleC+angleA)`
`=180^(@)-(110^(@)+20^(@))`
`180^(@)-130^(@)=50^(@)`
In `DeltaPBC, angleBPC+anglePCB+anglePBC=180^(@)`
[sum of all interion angles of any triangle is `180^(@)`]
`implies50^(@)+20^(@)+anglePBC=180^(@)`
`impliesanglePBC=180^(@)-70^(@)`
`impliesanglePBC=110^(@)`
Since, APB is a straight line.
`:.anglePBC+angleCBA=180^(@)`
`impliesangleCBA=180^(@)-110^(@)=70^(@)`