Answer :
Final answer:
To find ∠OQR, we use the properties of an isosceles triangle and the fact that a radius intersects with a tangent at a right angle. ∠OPQ and ∠OQP are each 45 degrees, making ∠OQR a right angle, i.e., 90 degrees.
Explanation:
To find ∠OQR in a circle where QP is a tangent and ΔOPQ is isosceles with center O at a point P on the circle, we must use geometric considerations and the properties of tangents to circles.
Firstly, since QP is a tangent and OP is a radius that intersects with the tangent at the point of tangency, we know that ∠OPQ is a right angle based on the property that a radius drawn to a point of tangency forms a right angle with the tangent line. Considering ΔOPQ to be isosceles and knowing that one angle is 90 degrees (right angle), the other two angles must be equal. In an isosceles triangle, the angles opposite the equal sides are equal. Since OP=OQ (given that ΔOPQ is isosceles), ∠OQP = ∠OPQ.
With the angle ∠OPQ being 90 degrees from the tangent property and ∠OQP being equal to ∠OPQ in an isosceles triangle, both ∠OPQ and ∠OQP are 45 degrees. Now, angle ∠OQR is the exterior angle for ΔOPQ, and according to the exterior angle theorem, ∠OQR = ∠OPQ + ∠OQP. Therefore, ∠OQR = 45 degrees + 45 degrees = 90 degrees. Angle ∠OQR, being formed by the tangent QP and the extension of the radius OQ, is also a right angle.
This conclusion doesn't require additional distances or the use of the Pythagorean theorem; it relies on the properties of isosceles triangles and circle tangents.
