Exercise 11.2 — Specific Angles

Exercise 11.2 — Trigonometric Ratios of Specific Angles

Exercise 11.2 is one of the most important exercises in Chapter 11, Trigonometry, for Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). It introduces the trigonometric ratios of the five standard angles — 0°, 30°, 45°, 60°, and 90° — and teaches you how to use them to evaluate tricky-looking expressions, verify identities, and solve right-triangle problems without ever needing a calculator.

Every value in this exercise is built from just two special triangles: a right isosceles triangle (which gives 45°) and an equilateral triangle cut in half (which gives 30° and 60°). Once those values are memorised, the rest of the exercise is simply substitution, careful fraction work, and applying the ratios to real right-triangle questions.

Deriving the Ratios of 45° — The Right Isosceles Triangle

Consider a right-angled triangle ABC where ∠B = 90° and the two legs are equal: AB = BC = a units. Because the triangle is isosceles, the two base angles must also be equal, so ∠A = ∠C = 45° each.

By the Pythagoras theorem, the hypotenuse AC can be found directly from the two equal legs:

Now the three primary ratios for ∠A = 45° follow straight from the triangle's sides:

The reciprocal ratios then follow automatically:

Deriving the Ratios of 30° and 60° — The Equilateral Triangle

Now take an equilateral triangle ABC with every side equal to 2a units, so every angle is 60°. Drop a perpendicular AD from vertex A to side BC. Because the triangle is equilateral, this perpendicular does two jobs at once: it bisects ∠A into two 30° angles, and it bisects the side BC into two equal halves of length a each.

Looking at the right triangle ABD alone: AB = 2a, BD = a, ∠B = 60°, ∠BAD = 30°, and ∠ADB = 90°. The third side AD is found using Pythagoras:

Ratios of 60° (using ∠B in triangle ABD)

Ratios of 30° (using ∠A in triangle ABD)

Ratios of 0° and 90° — The Limiting Cases

Picture a segment AC of fixed length r, making an acute angle θ with a horizontal ray AB, where BC is the perpendicular height of point C above the ray. As θ shrinks towards 0°, point C slides down until it almost lies on the ray itself — making BC nearly 0 and AB nearly equal to AC. As θ grows towards 90°, the opposite happens: AB shrinks to 0 while BC grows to match AC.

When θ = 0°

Here BC = 0 and AB = AC = r, so the ratios become:

When θ = 90°

Here AB = 0 and BC = AC = r, so the ratios flip around:

The Complete Table of Trigonometric Ratios

Combining every derivation above gives the single most useful table in the whole chapter. CBSE, Telangana, and Andhra Pradesh board exams draw questions from this table constantly — it is worth memorising completely before attempting Exercise 11.2 or any later exercise in this chapter.

Question 1 — Evaluate the Following Expressions

This question gives five expressions built entirely from the standard-angle table above. The method every time is the same: substitute the exact value for each ratio, then simplify the resulting fraction or surd expression carefully.

Question 2 — Choose the Right Answer and Justify

This question presents three expressions, each matching one of four multiple-choice options. The fastest approach is to substitute the standard values, simplify down to a single ratio, and then match that ratio against the table.

Question 3 — Testing sin 60°cos 30° + sin 30°cos 60° vs sin 90°

This question asks you to evaluate two separate expressions and compare the results, leading to one of the most important identities in trigonometry.

Question 4 — Is cos(60° + 30°) = cos 60° cos 30° − sin 60° sin 30°?

Question 5 — Finding QR and PR in a Right Triangle

In right triangle PQR, the right angle is at Q, with PQ = 6 cm and ∠RPQ = 60°. This is a direct application question: once one side and one acute angle of a right triangle are known, every other side can be found using the standard ratios.

Question 6 — Finding the Angles in Triangle XYZ

In right triangle XYZ, the right angle is at Y, with YZ = x and the hypotenuse XZ = 2x. Here the side lengths are given instead of an angle, so the strategy is reversed: build a ratio from the known sides, then match it against the standard-angle table to identify the angle itself.

Question 7 — Is sin(A + B) Equal to sin A + sin B?

This is a conceptual trap question. It is tempting to assume trigonometric ratios distribute over addition the way ordinary multiplication does, but this question proves that assumption false using A = 60° and B = 30°.

Common Mistakes to Avoid

• Treating trig functions as linear: sin(A+B), cos(A+B), and tan(A+B) can never be split into separate terms like sinA + sinB — always use the correct compound angle formula.
• Forgetting undefined values: tan 90°, cot 0°, cosec 0°, and sec 90° are undefined, not zero or infinity written as a number — write "not defined" in exam answers.
• Mixing up sin and cos rows: Since the sin and cos rows are mirror images of each other (0,1/2,1/√2,√3/2,1 vs. 1,√3/2,1/√2,1/2,0), it is easy to swap them by mistake under exam pressure.
• Not rationalising final answers: Expressions like 2/√2 should always be simplified to √2 by rationalising the denominator before writing the final answer.
• Squaring errors with surds: When squaring a value like √3/2, remember (√3/2)² = 3/4, not 3/2 — a very common silly mistake in Question 1, parts (iv) and (v).
• Skipping the angle-sum check: In triangle problems like Question 6, always verify that the three angles found actually add up to 180° before finalising the answer.

Quick Reference — All Answers at a Glance

What This Exercise Prepares You For

Exercise 11.2 is the computational backbone of the entire Trigonometry chapter. Once the standard-angle table and the compound angle ideas from Questions 3, 4, and 7 are solid, the next step is learning the fundamental trigonometric identities — sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ — which several parts of Question 1 already hint at.

The right-triangle skills practised in Questions 5 and 6 connect directly to heights and distances problems later in the chapter, where the same ratios are used to find unknown heights using angles of elevation and depression. For revision of the underlying triangle properties, see Triangles and the angle-sum property used in Question 6.