a circle having a centre O has a triangle PQS inside it with the base QP is not on centre O. PT and QT are two tangents from an exterior point T (S, O and T are linear points - lie in one line )
it is given that angle ptq = 25 deg . find angle PSQ and OPQ
Take a look at paragraph 5 of
http://www.regentsprep.org/Regents/math/geometry/GP15/CircleAngles.htm
To find the angles PSQ and OPQ, we can use the properties of tangents and the angles within a circle.
First, let's establish some properties:
1. Tangents drawn from an external point to a circle are equal in length.
This means that TP = TQ.
2. The measure of an angle formed by a chord and a tangent drawn from the same point of contact is equal to half of the measure of the arc intercepted by the angle at the center of the circle.
In other words, angle PTQ = 1/2 * arc PQ.
Now, let's apply these properties to solve the problem:
Since TP = TQ, triangle PTQ is an isosceles triangle. Thus, angle TPQ = angle TQP.
Using the property mentioned above (angle PTQ = 1/2 * arc PQ), we know that angle PTQ = 25 degrees. Therefore, arc PQ = 2 * angle PTQ = 50 degrees.
Because angles PTQ and PQT are equal, triangle PQS is also an isosceles triangle. Thus, angle PSQ = angle PQS.
We also know that angle PTQ = angle TPQ, and since angle TPQ and angle PTQ are both part of angle OPQ, we have that angle OPQ = 2 * angle TPQ = 2 * 25 degrees = 50 degrees.
To summarize:
- Angle PSQ = angle PQS.
- Angle OPQ = 50 degrees.
Please note that these values are applicable for the given conditions in the problem.