PRT is a tangent to a circle with centre O, at the point R on it. Diameter SQ of the circle is produced to meet the tangent at P and QR is jointed. If ∠QRP = 28°, then the measure of ∠SPR is:
34°
Solving Circle Geometry: Finding ∠SPR
This problem involves a circle, a tangent line, a diameter, and several angles. We are given the measure of one angle and asked to find another. We will use fundamental geometry theorems related to circles and triangles to solve this problem.
Problem Setup
We have a circle with centre O. PRT is a tangent touching the circle at point R. SQ is a diameter that is extended to meet the tangent at point P. QR is a chord joining Q and R. We are given that ∠QRP = 28°.
We need to find the measure of ∠SPR.
Step-by-Step Solution
Let's break down the problem and find the required angle using known geometric properties:
- Radius is Perpendicular to Tangent: The radius OR drawn to the point of contact R is perpendicular to the tangent line PRT. Therefore, ∠ORP = 90°.
- Finding ∠ORQ: We know ∠ORP is composed of ∠ORQ and ∠QRP. So, we can write:
\(\text{∠ORQ} = \text{∠ORP} - \text{∠QRP}\)
Substituting the known values:
\(\text{∠ORQ} = 90° - 28° = 62°\)
- Identifying Isosceles Triangle ▵ORQ: O is the centre of the circle, and OR and OQ are both radii. Thus, ▵ORQ is an isosceles triangle with OR = OQ.
- Angles in Isosceles Triangle: In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, ∠OQR = ∠ORQ.
\(\text{∠OQR} = 62°\)
- Relating ∠OQR to ∠SQR: Since S, Q, and P lie on a straight line and Q is between S and P, the angle ∠OQR is the same as ∠SQR.
\(\text{∠SQR} = 62°\)
- Angles on a Straight Line: The points S, Q, and P form a straight line. The angles ∠SQR and ∠RQP are supplementary because they form a linear pair (or angles on a straight line at point Q).
\(\text{∠RQP} + \text{∠SQR} = 180°\)
Substituting the value of ∠SQR:
\(\text{∠RQP} + 62° = 180°\)
\(\text{∠RQP} = 180° - 62° = 118°\)
- Sum of Angles in ▵PQR: Now consider the triangle ▵PQR. The sum of the interior angles in a triangle is 180°.
\(\text{∠QPR} + \text{∠PQR} + \text{∠QRP} = 180°\)
We know ∠PQR = ∠RQP = 118° and we are given ∠QRP = 28°. ∠QPR is the same as ∠SPR.
Substituting these values:
\(\text{∠SPR} + 118° + 28° = 180°\)
\(\text{∠SPR} + 146° = 180°\)
- Finding ∠SPR:
\(\text{∠SPR} = 180° - 146° = 34°\)
Summary of Angle Calculations
| Angle | Calculation/Reason | Measure |
|---|---|---|
| ∠ORP | Radius ⊥ Tangent | 90° |
| ∠ORQ | ∠ORP - ∠QRP | 90° - 28° = 62° |
| ∠OQR | Base angle of isosceles ▵ORQ (OR=OQ) | 62° |
| ∠SQR | Same as ∠OQR (S, Q, P collinear) | 62° |
| ∠RQP (or ∠PQR) | Supplementary to ∠SQR (angles on straight line SP) | 180° - 62° = 118° |
| ∠SPR (or ∠QPR) | Sum of angles in ▵PQR (180° - ∠PQR - ∠QRP) | 180° - 118° - 28° = 34° |
Thus, the measure of ∠SPR is 34°.
Revision Table: Circle Tangent Geometry
| Concept | Description | Application in this Problem |
|---|---|---|
| Tangent-Radius Property | A tangent to a circle is perpendicular to the radius at the point of contact. | Used to establish ∠ORP = 90°. |
| Isosceles Triangle Properties | In an isosceles triangle, angles opposite equal sides are equal. | Used in ▵ORQ (OR=OQ) to find ∠OQR from ∠ORQ. |
| Angles on a Straight Line | Angles forming a linear pair on a straight line sum to 180°. | Used to find ∠PQR from ∠SQR. |
| Sum of Angles in a Triangle | The sum of the interior angles of any triangle is 180°. | Used in ▵PQR to find ∠SPR. |
| Alternate Segment Theorem | The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. (Not strictly needed for this solution path, but applicable) | ∠QRP = ∠RSQ = 28°. (Could be used to find ∠RSQ, which is 28°, consistent with ▵SQR angles 90, 62, 28). |
Additional Information: Circle Properties
Understanding these properties is crucial for solving circle geometry problems:
- Diameter: A chord passing through the center of the circle. It is the longest chord.
- Tangent: A line that touches the circle at exactly one point (the point of contact).
- Chord: A line segment connecting two points on the circle.
- Angle in a Semicircle: The angle subtended by a diameter at any point on the circumference is 90°. In our diagram, ∠SRQ = 90° because SQ is a diameter.
- Angles Subtended by the Same Arc: Angles subtended by the same arc at the circumference are equal.
By applying these fundamental theorems, geometry problems involving circles and angles can be systematically solved.
Similar Questions
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Important Questions from Circles, Chords and Tangents
AB is a chord of a circle with centre O and P is any point on the circle. If ∠APB = 112°, then what is the measure of ∠OAB ?
The distance between the centres of two circles is 24 cm. If the radius of the two circles are 4 cm and 8 cm, then what is the sum of the lengths (in cm) of the direct common tangent and the transverse common tangent?
An equilateral triangle ABC and a scalene triangle DBC are inscribed in a circle on same side of the arc. what is ∠BDC equal to?
What can be the maximum number of common tangent which can be drawn to two non-intersecting circles?
There are two identical circles of radius 10 cm each. If the length of the direct common tangent is 26 cm, then what is the length (in cm) of the transverse common tangent?
