Class 10 Maths Chapter 9 Exercise 9.2 Solutions 2026 – Tangent and Angles of a Circle Notes PDF

Unit 9: Tangent and Angles of a Circle | Exercise 9.2 | Punjab Board New Syllabus 2026–27

Updated August 2026: Exercise 9.2 solutions for Tangent and Angles of a Circle have been uploaded below and are free to view or download, fully solved according to the new PCTB/PECTAA Class 10 Mathematics syllabus.

Angles Seen from Different Points on a Circle

The same arc of a circle can be “seen” from several different viewpoints — from the center, or from any point on the remaining part of the circle — and each viewpoint gives a specific, related angle. Exercise 9.2 introduces the theorems connecting these angles, building on the tangent properties from Exercise 9.1.

Key Concepts Covered in Exercise 9.2

The Central Angle Theorem

The angle that an arc subtends at the center of a circle is always exactly double the angle it subtends at any point on the remaining (major) part of the circle. This is the single most important theorem in this exercise.

Angle in a Semicircle

As a direct consequence of the central angle theorem, the angle subtended by a diameter at any point on the circle is always exactly 90°, since the central angle for a diameter is 180° (half of it is 90°).

Angles in the Same Segment

Angles subtended by the same arc, at different points within the same segment of the circle, are always equal to each other — this follows because they are all related to the same fixed central angle.

Applying These Theorems Together

Many questions in this exercise require combining two of these theorems — for example, using the angle-in-a-semicircle result inside a larger triangle, or applying the central angle theorem before comparing angles in the same segment.

Step-by-Step Solved Examples

Q. No. 1: An arc subtends an angle of 70° at a point on the remaining part of the circle. Find the angle it subtends at the center.

Central angle = 2 × angle at circumference = 2 × 70°

Central angle = 140°

Q. No. 2: AB is a diameter of a circle, and C is any point on the circle. Find ∠ACB.

Since AB is a diameter, it subtends a central angle of 180°.

By the angle-in-a-semicircle theorem, ∠ACB = 180°/2 = 90°

Q. No. 3: Points C and D lie in the same segment of a circle, on the same side of chord AB. If ∠ACB = 55°, find ∠ADB.

Since C and D are in the same segment and both angles are subtended by the same arc AB:

∠ADB = ∠ACB = 55° (angles in the same segment are equal)

These three questions are representative samples. The full PDF notes uploaded on this page walk through every question of Exercise 9.2 in the same numbered, step-by-step format.

MCQs, Short Questions & Long Questions from This Exercise

Alongside the main exercise, this page’s uploaded notes are organized the way Punjab Board papers are structured, so students can practice by question type:

• MCQs: Quick questions on applying the central angle theorem or identifying an angle in a semicircle
• Short Questions: Finding a central or circumference angle directly, like Q. No. 1 or 2 above
• Long Questions: Multi-step problems combining two or more angle theorems, like Q. No. 3 above within a larger figure

Where This Matters Beyond This Exercise

These angle theorems, combined with the tangent properties from Exercise 9.1, form the complete toolkit needed for the circle-geometry proof questions common in board exams. They also underpin the geometric constructions studied in the next chapter, Practical Geometry of Circles, where inscribed and circumscribed circles rely on these same angle relationships.

Common Mistakes Students Make in Exercise 9.2

• Forgetting to double (or halve) the angle correctly when switching between the center and the circumference
• Assuming any two angles in a circle are equal, without checking they’re in the same segment subtending the same arc
• Missing that a diameter is a special case that always produces a 90° angle at the circumference
• Confusing the major and minor segments when identifying which angles are equal

Why This Exercise Matters for the Board Exam

Angle-in-a-semicircle and same-segment questions are compact, elegant, and extremely common in the circle-geometry section of the board paper, often appearing as part of a larger multi-step proof. Recognizing which theorem applies at a glance, from a labeled diagram, is often the fastest route to a full-mark answer.

Quick Links – Chapter 9: Tangent and Angles of a Circle

Links for sections other than Exercise 9.2 will be activated as they are published on this site.

Download Exercise 9.2 Notes PDF

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Frequently Asked Questions (FAQs)

Q1. Are these Exercise 9.2 notes free to download?

Yes, all notes on this page are completely free to view and download in PDF format.

Q2. Which board are these notes for?

These notes are prepared according to the Punjab Board syllabus and are useful for all Punjab boards (Lahore, Gujranwala, Multan, Sargodha, Rawalpindi, Faisalabad, DG Khan, Bahawalpur, Sahiwal) as well as the Federal Board (FBISE).

Q3. Why is the angle in a semicircle always 90°?

A diameter subtends a straight angle (180°) at the center. By the central angle theorem, the angle at the circumference is always half the central angle, so 180° ÷ 2 = 90°, no matter where on the circle the point is chosen.

Q4. Does this page include MCQs and short questions too?

Yes, the uploaded notes include MCQs, short questions, and long questions related to this exercise’s topics, organized by question type to match the board paper pattern.

Q5. How can I download the PDF?

Click the “Download PDF” button above and the notes will open or download directly to your device.

Q6. Are these notes updated for the new 2026–27 syllabus?

Yes, these notes are prepared strictly according to the latest PCTB/PECTAA syllabus for the 2026–27 academic session.

Comments & Feedback

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