Trigonometry Numericals: A Practice Sequence From Easy to Board-Level

Most students do not struggle with trigonometry because it is hard. They struggle because they practise it in the wrong order. A student who jumps from a textbook exercise on ratios straight to a five-mark heights and distances problem hits a wall, blames the chapter, and starts skipping questions. The fix is not more hours. It is a better sequence.

This guide lays out that sequence. Each level builds on the one before it, with a worked example at every step so you can see what a correct answer looks like. If you have not yet read the chapter itself, start with our Introduction to Trigonometry page or the chapter 8 notes, then come back here to practise.

Why a sequence works better than random practice

Trigonometry in Class 10 is really three skills stacked on top of each other. First, you read a right triangle and name its sides correctly. Second, you use ratios, standard angle values and identities to manipulate expressions. Third, you turn a word problem into a diagram and apply the first two skills inside it.

Start here: reading ratios off a right triangle

Before any formula, practise one habit: labelling the sides from the angle you are working with. The hypotenuse never changes. The opposite and adjacent sides swap depending on which acute angle you pick.

Take a triangle ABC with the right angle at B, AB = 3 cm, BC = 4 cm and AC = 5 cm. For angle A:

• sin A = opposite / hypotenuse = BC / AC = 4/5
• cos A = adjacent / hypotenuse = AB / AC = 3/5
• tan A = opposite / adjacent = BC / AB = 4/3

Now switch to angle C. The opposite side becomes AB and the adjacent side becomes BC, so sin C = 3/5 and cos C = 4/5. If that swap feels natural, you are ready to move on. If not, redraw five different triangles and label them for both acute angles before you continue. The trigonometric ratios page has more labelled examples if you want to compare your work.

The next skill in this level is the reverse problem: you are given one ratio and must find the others. Suppose sin θ = 5/13. Draw a right triangle with opposite side 5 and hypotenuse 13. The adjacent side comes from Pythagoras: 13² − 5² = 169 − 25 = 144, so the adjacent side is 12. That gives cos θ = 12/13 and tan θ = 5/12. A quick refresher on the Pythagoras theorem helps here, because most slips in this type come from a small arithmetic error in the third side.

Next: the standard angle table, built instead of crammed

The values for 0°, 30°, 45°, 60° and 90° appear in almost every trigonometry question in Class 10, and students who cannot recall them instantly lose time on every single problem. You can memorise the table, but a pattern is easier to keep.

Write sin for the angles 0°, 30°, 45°, 60°, 90° as the square root of 0, 1, 2, 3, 4 divided by 2. That gives 0, 1/2, √2/2 (which is 1/√2), √3/2 and 1. Cosine is the same list in reverse order. Tangent is sin divided by cos, which gives 0, 1/√3, 1, √3 and not defined at 90°.

• Evaluate sin 30° cos 60° + sin 60° cos 30°. This is (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4 = 1.
• Evaluate 2 tan² 45° + cos² 30° − sin² 60°. This is 2(1) + 3/4 − 3/4 = 2.

The second answer does not depend on the cosine and sine terms at all, because they cancel. Our page on the trigonometric ratios of some specific angles lists the full table, and the trigonometry formulas sheet is worth printing and keeping beside your desk for the first week.

Then: finding unknown sides from one angle and one side

This is the bridge between pure algebra and the word problems you will meet later. You are given a right triangle, one acute angle and one side, and you must find another side.

Example: in triangle ABC, right-angled at B, AC = 10 cm and angle A = 30°. Find BC and AB.

• BC is opposite to angle A and AC is the hypotenuse, so sin 30° = BC/10. Since sin 30° = 1/2, BC = 5 cm.
• AB is adjacent to angle A, so cos 30° = AB/10. Since cos 30° = √3/2, AB = 5√3 cm, which is about 8.66 cm.

Level up: identities, where most proof marks are won or lost

Identities scare students because they look like a different subject. They are not. Three facts do almost all the work: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. Our trigonometric identities page shows where they come from, and the memory framework for trigonometric identities article explains how to keep them without rote repetition.

The most reliable strategy for a proof is to convert everything to sin and cos, then simplify one side until it matches the other. Here is a standard example.

Prove that sec θ − cos θ = sin θ tan θ.

• Start with the left side: sec θ − cos θ = 1/cos θ − cos θ.
• Take a common denominator: (1 − cos²θ) / cos θ.
• Use the first identity: 1 − cos²θ = sin²θ, so the expression becomes sin²θ / cos θ.
• Split it as sin θ × (sin θ / cos θ) = sin θ tan θ, which is the right side.

Write every step in a proof. Boards award marks for the steps shown, so a correct line of working earns credit even when the last line slips. For a deeper bank of questions, work through the Class 10 trigonometry important questions page once you have finished the examples here.

Board-level: heights and distances from a single diagram

This is the stage the earlier ones were preparing you for. Read the chapter on some applications of trigonometry first if the vocabulary of angle of elevation and angle of depression is new. The notes in applications of trigonometry, quick revision summarise it on a single page.

Example one, a single angle: the angle of elevation of the top of a tower from a point 30 m from its base is 60°. Find the height of the tower.

• The height is opposite to 60° and the 30 m distance is adjacent, so tan 60° = h / 30.
• Since tan 60° = √3, h = 30√3 m.
• Using √3 = 1.732, h = 51.96 m, which is about 52 m.

Example two, an angle of depression: from the top of a 50 m tall building, the angle of depression of a car on the road is 30°. How far is the car from the foot of the building?

• The angle of depression equals the angle of elevation from the car to the top, by alternate angles, so the angle at the car is 30°.
• The 50 m height is opposite and the required distance d is adjacent, so tan 30° = 50 / d.
• Since tan 30° = 1/√3, d = 50√3 m, which is about 86.6 m.

The hardest version asks for two observations of the same object, such as the angle of elevation from two points on a line. Here you write two equations that share the height, then solve for the unknown distance. Practise this type last, and only after single-angle questions feel routine. Our article on common diagram mistakes in heights and distances covers the slips that cost marks even when the math is correct.

A two-week plan for working through the sequence

A plan removes the daily question of what to practise next. Use this as a starting point and stretch it if you need more time on a stage. The idea is to finish trigonometry questions for Class 10 in the order above, with a short review every few days.

Slips that quietly cost marks

Most lost marks in this chapter come from a handful of repeat errors, not from missing knowledge.

• Labelling sides from the wrong angle. After you switch to the other acute angle, opposite and adjacent swap. Re-label every time.
• Rounding √3 too early. Carry √3 through the working and substitute 1.732 only in the last line, unless the question says otherwise.
• Skipping the figure. A word problem solved without a diagram is the main source of wrong ratios in heights and distances.
• Mixing up sec and cosec. Sec goes with cos and cosec goes with sin. Say it aloud once and it sticks.
• Jumping steps in a proof. Writing sin²θ + cos²θ = 1 as a separate line costs you nothing and often earns a mark.
• Not rationalising. If your answer ends with 1/√3 in the denominator, check whether the question expects a rationalised form.

Frequently Asked Questions

Q1. How many trigonometry questions should I solve every day?

Six to eight is enough for most students in the early stages, and four to five for the harder stages such as proofs and heights and distances. Quality matters more than quantity. Review every wrong answer the same day, because one corrected mistake teaches more than five new questions solved on autopilot.

Q2. Should I memorise the trigonometric table or derive it?

Do both. Learn the pattern of square roots of 0 to 4 divided by 2 so you can rebuild the sine row in seconds, then practise until the values come without thinking. During the exam, writing the row quickly in the margin is a reasonable way to avoid a recall error.

Q3. How do I know which identity to use in a proof?

Start by converting everything to sin and cos. If you see 1 minus a squared ratio, think of the first identity. If you see 1 plus tan squared or 1 plus cot squared, think of the other two. With practice, the pattern in the expression points you to the identity.

Q4. Are heights and distances problems always asked in the board exam?

They are a regular part of the CBSE and Karnataka SSLC papers, though the exact question types and marks can change from year to year. Check the latest blueprint and sample papers from your own board to see how many marks they carry, and then give the topic enough practice time.

Q5. How can I check my answer to a trigonometry problem?

Three quick checks help. First, the hypotenuse must be the longest side. Second, sin and cos of an acute angle must lie between 0 and 1. Third, for identity proofs, test the claim with a simple angle such as 30° or 45° before you start. If both sides disagree there, you have misread the question.

Conclusion: build trigonometry up one step at a time

Trigonometry rewards order. Learn to label a right triangle, make the standard values automatic, use them to find sides, prove identities by converting to sin and cos, and only then take on heights and distances. Two weeks of focused practice in that sequence will do more than a month of mixed, unplanned questions. Keep a short list of your own repeat mistakes and read it before every practice session.