Class 10 Maths Trigonometry: 40 Important Questions with Step-by-Step Solutions

Trigonometry in Class 10 clusters into a small number of recurring question types — once you've seen all of them mapped out, "important questions" stops feeling like a vague label and starts looking like a checklist. This is 40 original problems, 5 for every distinct type, each with a complete solution you can follow line by line.

How to use this page: Read each problem, attempt it fully on paper first, then tap "Reveal Solution" to check your working — not just your final answer. Every single calculation on this page was computed and independently verified before publishing.

Every Question Type Covered

Finding all trig ratios from one given ratio

Evaluating expressions using standard angle values

Using complementary angle relationships

Proving trigonometric identities

Simplifying trigonometric expressions

Heights & Distances — single angle of elevation

Heights & Distances — two angles or two positions

Heights & Distances — angle of depression

Quick reference — standard angle values used throughout:

Angle	0°	30°	45°	60°	90°
sin	0	1/2	1/√2	√3/2	1
cos	1	√3/2	1/√2	1/2	0
tan	0	1/√3	1	√3	undefined

Type 1 · Finding Trig Ratios

1.1

Given sinA

If sinA = 3/5, find cosA and tanA.

Reveal Solution

Set up the triangle

sinA = opposite/hypotenuse = 3/5. Let opposite = 3, hypotenuse = 5.

adjacent² = hyp² − opp² = 5² − 3² = 16 → adjacent = 4

✓ cosA = 4/5, tanA = 3/4

1.2

Given cosA

If cosA = 12/13, find sinA and tanA.

Reveal Solution

Set up the triangle

cosA = adjacent/hypotenuse = 12/13.

opposite² = 13² − 12² = 25 → opposite = 5

✓ sinA = 5/13, tanA = 5/12

1.3

Given tanA

If tanA = 4/3, find sinA and cosA.

Reveal Solution

Set up the triangle

tanA = opposite/adjacent = 4/3.

hypotenuse² = 4² + 3² = 25 → hypotenuse = 5

✓ sinA = 4/5, cosA = 3/5

⚠️ Common mistake: confusing tanA's numerator/denominator with sinA's — tan uses opposite/adjacent, not opposite/hypotenuse.

1.4

Given cotA

If cotA = 15/8, find sinA and cosA.

Reveal Solution

Set up the triangle

cotA = adjacent/opposite = 15/8, so adjacent = 15, opposite = 8.

hypotenuse² = 15² + 8² = 289 → hypotenuse = 17

✓ sinA = 8/17, cosA = 15/17

1.5

Given secA

If secA = 25/24, find sinA and tanA.

Reveal Solution

Set up the triangle

secA = hypotenuse/adjacent = 25/24, so hypotenuse = 25, adjacent = 24.

opposite² = 25² − 24² = 49 → opposite = 7

✓ sinA = 7/25, tanA = 7/24

Type 2 · Standard Angle Expressions

2.1

Evaluate

Evaluate: sin30°cos60° + cos30°sin60°

Reveal Solution

(1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4

✓ = 1

2.2

Evaluate

Evaluate: 2cos²45° − 1

Reveal Solution

2 × (1/√2)² − 1 = 2 × (1/2) − 1 = 1 − 1

✓ = 0

2.3

Evaluate

Evaluate: sin²30° + cos²30°

Reveal Solution

(1/2)² + (√3/2)² = 1/4 + 3/4

✓ = 1 (confirms the Pythagorean identity at this specific angle)

2.4

Evaluate

Evaluate: tan60°/tan30°

Reveal Solution

√3 ÷ (1/√3) = √3 × √3

✓ = 3

2.5

Evaluate

Evaluate: cos60°cos30° − sin60°sin30°

Reveal Solution

(1/2)(√3/2) − (√3/2)(1/2) = √3/4 − √3/4

✓ = 0

Type 3 · Complementary Angles

3.1

Complementary Ratio

Evaluate: sin65° / cos25°

Reveal Solution

Key relation

Since 25° = 90° − 65°, cos25° = cos(90°−65°) = sin65°.

sin65° / sin65° = 1

✓ = 1

3.2

Complementary Ratio

Evaluate: tan48° × tan42°

Reveal Solution

Key relation

42° = 90° − 48°, so tan42° = cot48°.

tan48° × cot48° = 1

✓ = 1

3.3

Complementary Ratio

Evaluate: cosec31° / sec59°

Reveal Solution

Key relation

59° = 90° − 31°, so sec59° = cosec31°.

cosec31° / cosec31° = 1

✓ = 1

3.4

Complementary Ratio

Evaluate: tan15° × tan75°

Reveal Solution

Key relation

75° = 90° − 15°, so tan75° = cot15°.

tan15° × cot15° = 1

✓ = 1

⚠️ Common mistake: trying to evaluate tan15° and tan75° individually using half-angle formulas — spotting the complementary pair first avoids that entirely.

3.5

Complementary Ratio

Evaluate: sin²35° + sin²55°

Reveal Solution

Key relation

55° = 90° − 35°, so sin55° = cos35°.

sin²35° + cos²35° = 1 (Pythagorean identity)

✓ = 1

Type 4 · Proving Identities

4.1

Prove

Prove that: (secθ − tanθ)(secθ + tanθ) = 1

Reveal Solution

Expand as a difference of squares

LHS = sec²θ − tan²θ

Using the identity 1 + tan²θ = sec²θ, we get sec²θ − tan²θ = 1

✓ LHS = 1 = RHS. Proved.

4.2

Prove

Prove that: sinθ/(1+cosθ) + (1+cosθ)/sinθ = 2cosecθ

Reveal Solution

Combine over a common denominator

LHS = [sin²θ + (1+cosθ)²] / [sinθ(1+cosθ)]

Expand the numerator

sin²θ + 1 + 2cosθ + cos²θ = (sin²θ+cos²θ) + 1 + 2cosθ = 1 + 1 + 2cosθ = 2 + 2cosθ = 2(1+cosθ)

LHS = 2(1+cosθ) / [sinθ(1+cosθ)] = 2/sinθ = 2cosecθ

✓ LHS = 2cosecθ = RHS. Proved.

4.3

Prove

Prove that: (1+tan²θ)/(1+cot²θ) = tan²θ

Reveal Solution

Use both Pythagorean identities

1+tan²θ = sec²θ, and 1+cot²θ = cosec²θ

LHS = sec²θ/cosec²θ = (1/cos²θ)/(1/sin²θ) = sin²θ/cos²θ = tan²θ

✓ LHS = tan²θ = RHS. Proved.

4.4

Prove

Prove that: cosθ/(1−tanθ) + sinθ/(1−cotθ) = sinθ + cosθ

Reveal Solution

Rewrite tanθ and cotθ in terms of sin and cos

First term: cosθ/(1−sinθ/cosθ) = cos²θ/(cosθ−sinθ)

Second term

sinθ/(1−cosθ/sinθ) = sin²θ/(sinθ−cosθ) = −sin²θ/(cosθ−sinθ)

                                LHS = [cos²θ − sin²θ]/(cosθ−sinθ) = (cosθ−sinθ)(cosθ+sinθ)/(cosθ−sinθ) = cosθ+sinθ
                            

✓ LHS = sinθ+cosθ = RHS. Proved.

⚠️ Common mistake: sign errors when combining the two fractions — carefully track that the second term's denominator becomes negative of the first's.

4.5

Prove

Prove that: (cosecθ − sinθ)(secθ − cosθ)(tanθ + cotθ) = 1

Reveal Solution

Simplify each bracket separately

cosecθ−sinθ = 1/sinθ − sinθ = (1−sin²θ)/sinθ = cos²θ/sinθ

secθ−cosθ = 1/cosθ − cosθ = (1−cos²θ)/cosθ = sin²θ/cosθ

tanθ+cotθ = sinθ/cosθ + cosθ/sinθ = (sin²θ+cos²θ)/(sinθcosθ) = 1/(sinθcosθ)

                                LHS = (cos²θ/sinθ) × (sin²θ/cosθ) × 1/(sinθcosθ) = (cos²θ·sin²θ)/(sinθ·cosθ·sinθ·cosθ)
                            

✓ = 1 = RHS. Proved.

Type 5 · Simplifying Expressions

5.1

Simplify

Simplify: sin⁴θ − cos⁴θ

Reveal Solution

Factor as a difference of squares

sin⁴θ−cos⁴θ = (sin²θ−cos²θ)(sin²θ+cos²θ)

Since sin²θ+cos²θ = 1, this reduces directly

✓ = sin²θ − cos²θ

5.2

Simplify

Simplify: (1 − cos²θ) × cosec²θ

Reveal Solution

Use the Pythagorean identity

1−cos²θ = sin²θ

sin²θ × cosec²θ = sin²θ × (1/sin²θ)

✓ = 1

5.3

Prove

Prove that: tanθ + cotθ = secθ × cosecθ

Reveal Solution

Write in terms of sin and cos

LHS = sinθ/cosθ + cosθ/sinθ = (sin²θ+cos²θ)/(sinθcosθ) = 1/(sinθcosθ)

RHS = secθ×cosecθ = (1/cosθ)×(1/sinθ) = 1/(sinθcosθ)

✓ LHS = RHS. Proved.

5.4

Prove

Prove that: (sinθ + cosecθ)² + (cosθ + secθ)² = 7 + tan²θ + cot²θ

Reveal Solution

Expand both squares

= sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

Group terms

= (sin²θ+cos²θ) + 4 + cosec²θ + sec²θ = 1 + 4 + (1+cot²θ) + (1+tan²θ)

= 5 + 1 + cot²θ + 1 + tan²θ = 7 + tan²θ + cot²θ

✓ LHS = RHS. Proved.

5.5

Prove

Prove that: √[(1+sinθ)/(1−sinθ)] = secθ + tanθ

Reveal Solution

Multiply inside the root by (1+sinθ)/(1+sinθ)

= √[(1+sinθ)²/((1−sinθ)(1+sinθ))] = √[(1+sinθ)²/(1−sin²θ)]

= √[(1+sinθ)²/cos²θ] = (1+sinθ)/cosθ = 1/cosθ + sinθ/cosθ

✓ = secθ + tanθ. Proved.

Type 6 · Heights & Distances — Single Angle

6.1

Elevation

The distance from the foot of a tower to a point on the ground is 20√3 m. If the angle of elevation of the top of the tower from this point is 30°, find the height of the tower.

Reveal Solution

tan30° = height/base → height = 20√3 × tan30° = 20√3 × (1/√3)

✓ Height = 20 m

6.2

Elevation

A point is 15√3 m from the base of a tower. The angle of elevation to the top of the tower is 60°. Find the tower's height.

Reveal Solution

height = 15√3 × tan60° = 15√3 × √3

✓ Height = 45 m

6.3

Elevation → Shadow

A pole is 10 m tall. When the sun's angle of elevation is 45°, find the length of the pole's shadow.

Reveal Solution

tan45° = height/shadow → shadow = height/tan45° = 10/1

✓ Shadow length = 10 m

6.4

Elevation

The foot of a ladder leaning against a wall is 8√3 m from the wall, making a 30° angle of elevation with the ground. Find how high up the wall the ladder reaches.

Reveal Solution

height = 8√3 × tan30° = 8√3 × (1/√3)

✓ Height on wall = 8 m

6.5

Elevation → Distance

A tower is 30 m tall. Find the distance of a point on the ground from the base of the tower, if the angle of elevation of the top from that point is 60°.

Reveal Solution

distance = height/tan60° = 30/√3 = 30/√3 × (√3/√3)

✓ Distance = 10√3 m

Type 7 · Heights & Distances — Two Angles

7.1

Two Positions

From a point on the ground, the angle of elevation to the top of a tower is 30°. Moving 20 m closer, the angle becomes 60°. Find the height of the tower.

Reveal Solution

Let height = h, near distance = d

From the near point: tan60° = h/d → d = h/√3

From the far point: tan30° = h/(d+20) → d+20 = h√3

Subtracting: h√3 − h/√3 = 20 → h(3−1)/√3 = 20 → h = 10√3

✓ Height of tower = 10√3 m ≈ 17.32 m

7.2

Two Positions

From a point, the angle of elevation to a tower's top is 30°. From a point 10(√3−1) m closer, it's 45°. Find the tower's height.

Reveal Solution

Set up both equations

Near: tan45° = h/d → d = h. Far: tan30° = h/(d+10(√3−1)) → d+10(√3−1) = h√3

Substituting d=h: h + 10(√3−1) = h√3 → 10(√3−1) = h(√3−1)

✓ Height of tower = 10 m

7.3

Two Positions

The angle of elevation to a tower's top is 45° from one point and 60° from a point 20 m closer. Find the tower's height.

Reveal Solution

Set up both equations

Near: tan60° = h/d → d = h/√3. Far: tan45° = h/(d+20) → d+20 = h

Substituting: h/√3 + 20 = h → 20 = h(1−1/√3) = h(√3−1)/√3

✓ Height of tower = 10√3 + 30 m ≈ 47.32 m

⚠️ This one doesn't simplify to as clean a number as the others — that's realistic. Not every heights-and-distances answer is a whole number; rationalising the denominator carefully still matters even when the final form has a surd in it.

7.4

Building & Tower

From the top of a 20 m building, the angle of elevation to the top of a nearby tower is 60°, and the angle of depression to its base is 45°. Find the height of the tower and the distance between the building and tower.

Reveal Solution

Find the horizontal distance first, using the depression angle

tan45° = 20/d → d = 20 m

Find the extra height above the building, using the elevation angle

tan60° = extra/d → extra = 20 × √3 = 20√3 m

Total tower height = building height + extra = 20 + 20√3

✓ Tower height = 20 + 20√3 m ≈ 54.64 m; distance between them = 20 m

7.5

Two Towers

Two towers stand on either side of a 30 m wide road. From the base of one tower, the angle of elevation to the top of the other is 60°, and from the base of the second, the angle of elevation to the top of the first is 30°. Find both heights.

Reveal Solution

Each tower forms its own right triangle with the full road width as base

Tower A: tan30° = height_A/30 → height_A = 30 × (1/√3)

Tower B: tan60° = height_B/30 → height_B = 30 × √3

                    ✓ Height of Tower A = 10√3 m ≈ 17.32 m; Height of Tower B = 30√3 m ≈ 51.96 m

Type 8 · Heights & Distances — Angle of Depression

8.1

Depression

From the top of a 60 m cliff, the angle of depression to a boat is 30°. Find the boat's distance from the base of the cliff.

Reveal Solution

The angle of depression equals the angle of elevation from the boat (alternate angles)

tan30° = 60/distance

distance = 60/tan30° = 60 × √3

✓ Distance = 60√3 m ≈ 103.92 m

8.2

Depression

From the top of a 75√3 m lighthouse, the angle of depression to a ship is 45°. Find the ship's distance from the lighthouse.

Reveal Solution

distance = 75√3/tan45° = 75√3/1

✓ Distance = 75√3 m ≈ 129.9 m

8.3

Depression — Two Boats

From the top of a 100 m lighthouse, the angles of depression to two boats in line with its base are 30° and 45°. Find the distance between the two boats.

Reveal Solution

Find each boat's distance separately

Far boat (30°): distance = 100/tan30° = 100√3. Near boat (45°): distance = 100/tan45° = 100

Distance between boats = 100√3 − 100

✓ Distance between boats = 100(√3−1) m ≈ 73.21 m

8.4

Depression — Pole & Tower

From the top of a tower, the angle of depression to the foot of a 10 m pole standing at some distance is 60°, and to the top of the same pole is 30°. Find the height of the tower.

Reveal Solution

Let tower height = H, horizontal distance = d

To pole foot: tan60° = H/d. To pole top: tan30° = (H−10)/d

Divide the two heights by d and subtract

H − (H−10) = d(tan60° − tan30°) → 10 = d(√3 − 1/√3) = d(2/√3)

d = 5√3, then H = d × tan60° = 5√3 × √3

✓ Height of tower = 15 m

⚠️ Common mistake: using the same angle for both the pole's foot and top — they're genuinely different lines of sight from the same tower-top point, giving two different angles.

8.5

Depression + Elevation

From a window 15 m above the ground, the angle of elevation to the top of a building across the road is 30°, and the angle of depression to its base is 45°. Find the height of the building.

Reveal Solution

Find the horizontal distance using the depression angle

tan45° = 15/d → d = 15 m

Find the extra height above the window using the elevation angle

tan30° = extra/d → extra = 15 × (1/√3) = 5√3

Total building height = window height + extra = 15 + 5√3

✓ Height of building = 15 + 5√3 m ≈ 23.66 m

Which of These 8 Types Would You Actually Recognise Without the Label?

Reading 40 solved problems and being able to solve fresh ones without the type named for you are different skills. The real exam test is spotting which of these 8 patterns a new question belongs to on sight.

What a Genelis weak area map looks like after working through Trigonometry practice sets

Finding ratios & standard angles

84%

Identity proofs & simplification

66%

Heights & distances — single angle

53%

Heights & distances — two angles

31%

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Frequently Asked Questions

Questions Students Commonly Ask

Quick answers to the most common questions related to this guide.

What question types appear in CBSE Class 10 Trigonometry?

Eight distinct types recur across CBSE Class 10 Trigonometry: finding all trigonometric ratios when one is given, evaluating expressions using standard angle values (0°, 30°, 45°, 60°, 90°), using complementary angle relationships, proving trigonometric identities, simplifying trigonometric expressions, and three variations of Heights and Distances — a single angle of elevation, two angles from two positions, and angle of depression problems.

What is the fastest way to find all trigonometric ratios when only one is given?

Draw a right triangle and label the given ratio's numerator and denominator as two of the three sides (for example, if sinA = 3/5, label the opposite side 3 and the hypotenuse 5). Use the Pythagorean theorem to find the third side, then read off every other ratio directly from the triangle. This is faster and less error-prone than manipulating identities algebraically for this specific question type.

What is the difference between angle of elevation and angle of depression?

Angle of elevation is measured upward from the horizontal, from an observer looking up at an object above them. Angle of depression is measured downward from the horizontal, from an observer looking down at an object below them. Both angles are measured from a horizontal line, not from the ground or from vertical — a common source of error is measuring from the wrong reference line entirely.

How do I solve a heights and distances problem involving two different angles from two points?

Set up two separate right-triangle equations using tan(angle) = height/distance for each of the two points, then solve them as simultaneous equations. Since the height is the same in both equations, express both distances in terms of height using cotangent, and use the known difference between the two distances to solve for height directly, typically using the relation height = (distance between the two points) / (cot of the smaller angle − cot of the larger angle).

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