The tangent drawn at points P and Q on the circumference of a circle intersect at A. If `angle PAQ =60^(@), ` find the value of `angle APQ.`

Explanation

Let P and Q be any two points on the circle with centre O. Two tangents AP and AQ drawn at the points P and Q intersect each other at the point A. Let us join P, Q and O, A
Given that `angle PAQ =60^(@).`
Now, Ap is a tangent to the circle with centre O and OP is the radius passing through the point of contact `P. therefore OP bot AP, therefore angle OPA=90^(@).`
Similarly, `angle OQA=90^(@)`
In `Delta's POA and Delta QOA,`
`angle OPA = angle OQA[because ` each is a right-angle]
`OP =OQ [because` radius of same circle] and hypotenuse OA is common to both
`therefore Delta POA -=Delta QOA` [by the R-H-S condition of congruency ]
`therefero angle OAP=angle OAQ` [similar angles of congurent triangles ]
Given that `angle PAQ =60^(@)or, angle PAO+ angle QAO=60^(@)`
or, `angle PAO =60^(@)angle QAO=angle PAO]`
or, `2 angle PAO=60^(@)or, angle PAO=(60^(@))/(2)=30^(@)`
Then, `angle PAO 90^(@)-angle OAP=90^(@)-30^(@) =60^(@)[becauseangleOAP=anglePAO=30^(@)]`
`therefore angleAPQ=angleAPO-angleOPQ=90^(@)-30^(@)=60^(@)`
`therefore ` value of `angle APQ=60^(@).`