Question 1

A person standing on the ground observes the angle of elevation to the top of a tree to be 45°. After walking 10 m towards the tree, the angle of elevation becomes 60°. Find the height of the tree.

Accepted Answer

Setup: Two-position angle of elevation problem using tangent ratios from two different distances.

Let h = height of the tree.
Let d = initial distance from the person to the base of the tree.

From the initial position:
 tan(45°) = h/d
 1 = h/d
 d = h ... (1)

After walking 10 m towards the tree, the distance becomes (d − 10):
 tan(60°) = h/(d − 10)
 √3 = h/(d − 10)
 d − 10 = h/√3 ... (2)

Substituting (1) into (2):
 h − 10 = h/√3
 h − h/√3 = 10
 h(1 − 1/√3) = 10
 h(√3 − 1)/√3 = 10
 h = 10√3/(√3 − 1)

Rationalize the denominator:
 h = 10√3/(√3 − 1) · (√3 + 1)/(√3 + 1)
 h = 10√3(√3 + 1)/(3 − 1)
 h = 10√3(√3 + 1)/2
 h = 10(3 + √3)/2
 h = 5(3 + √3)
 h = 15 + 5√3 metres

Verification: d = h = 15 + 5√3. After walking 10 m: new distance = 5 + 5√3. tan(60°) = (15 + 5√3)/(5 + 5√3) = 5(3 + √3)/(5(1 + √3)) = (3 + √3)/(1 + √3). Rationalize: [(3 + √3)(1 − √3)]/[(1 + √3)(1 − √3)] = [3 − 3√3 + √3 − 3]/[1 − 3] = [−2√3]/[−2] = √3 ✓

Question 2

A man standing at a point P observes the angle of elevation to the top of a vertical tower as 30°. He walks 100 m towards the base of the tower along level ground, and the angle of elevation becomes 60°. Find the height of the tower.

Accepted Answer

Setup: Use the tangent ratio in right triangles formed at two different observation points.

Let h = height of tower, and let the initial distance from P to the base be d metres.

From point P: tan(30°) = h/d
⟹ 1/√3 = h/d
⟹ d = h√3 ... (1)

After walking 100 m closer, distance from tower = (d − 100)
From new position: tan(60°) = h/(d − 100)
⟹ √3 = h/(d − 100)
⟹ d − 100 = h/√3 ... (2)

Substitute (1) into (2):
h√3 − 100 = h/√3
⟹ h√3 − h/√3 = 100
⟹ h(√3 − 1/√3) = 100
⟹ h(3/√3 − 1/√3) = 100
⟹ h(2/√3) = 100
⟹ h = 100√3/2 = 50√3 m

Verification: d = 50√3 · √3 = 150 m. New distance = 50 m. tan(60°) = 50√3/50 = √3 ✓

Question 3

From the top of a cliff 80 m high, the angle of depression to a boat on the water is 30°. How far is the boat from the base of the cliff (measured horizontally)?

Accepted Answer

Setup: Angle of depression from top of cliff equals angle of elevation from boat to top of cliff (alternate angles with horizontal).

Let x = horizontal distance from base of cliff to boat.
Height of cliff = 80 m.
Angle of depression = 30°.

From the top of the cliff, looking down at angle 30° below horizontal:
tan(30°) = (opposite)/(adjacent) = (height)/(horizontal distance)
⟹ tan(30°) = 80/x
⟹ 1/√3 = 80/x
⟹ x = 80√3 m

Verification: tan(30°) = 1/√3 ≈ 0.577, and 80/(80√3) = 1/√3 ✓

Question 4

Two vertical poles of heights 10 m and 15 m stand on level ground 12 m apart. Find the angle of elevation from the top of the shorter pole to the top of the taller pole.

Accepted Answer

Setup: Construct a right triangle where the vertical difference in heights forms one leg and the horizontal separation forms the other leg.

Height of shorter pole = 10 m
Height of taller pole = 15 m
Horizontal distance between poles = 12 m

Vertical difference in heights = 15 − 10 = 5 m

From the top of the shorter pole, looking up to the top of the taller pole:
tan(θ) = (vertical difference)/(horizontal distance)
⟹ tan(θ) = 5/12

Therefore, θ = arctan(5/12) = arctan(5/12)

To find the angle, note that tan(θ) = 5/12.
Using a calculator or recognizing the 5-12-13 Pythagorean triple:
sin(θ) = 5/13, cos(θ) = 12/13

θ ≈ 22.62° or arctan(5/12)

Exact answer: θ = arctan(5/12) ≈ 22.62°

Question 5

A ladder leans against a vertical wall. The angle between the ladder and the ground is 60°. If the ladder is 10 m long, find the height at which the top of the ladder touches the wall.

Accepted Answer

Setup: Right triangle with the ladder as the hypotenuse, the wall as the vertical side (opposite to the 60° angle), and the ground as the horizontal side (adjacent to the 60° angle).

Length of ladder (hypotenuse) = 10 m
Angle between ladder and ground = 60°
Height on wall = h (opposite side to the 60° angle)

Using sine ratio:
sin(60°) = (opposite)/(hypotenuse) = h/10
⟹ √3/2 = h/10
⟹ h = 10 · √3/2 = 5√3 m

Verification: sin(60°) = √3/2, so h = 10 · √3/2 = 5√3 ≈ 8.66 m ✓
Check: If h = 5√3, then base = 10·cos(60°) = 10·(1/2) = 5 m.
Pythagorean check: (5√3)² + 5² = 75 + 25 = 100 = 10² ✓

Question 6

From a point on level ground, the angle of elevation to the top of a building is 45°. After walking 50 m towards the building, the angle of elevation becomes 60°. Find the height of the building.

Accepted Answer

Setup: Use tangent ratios from two observation points to set up simultaneous equations.

Let h = height of building, and let initial distance from building be d metres.

From initial point: tan(45°) = h/d
⟹ 1 = h/d
⟹ d = h ... (1)

After walking 50 m closer, distance from building = (d − 50)
From new position: tan(60°) = h/(d − 50)
⟹ √3 = h/(d − 50)
⟹ d − 50 = h/√3 ... (2)

Substitute (1) into (2):
h − 50 = h/√3
⟹ h − h/√3 = 50
⟹ h(1 − 1/√3) = 50
⟹ h(√3 − 1)/√3 = 50
⟹ h = 50√3/(√3 − 1)

Rationalize denominator:
h = 50√3/(√3 − 1) · (√3 + 1)/(√3 + 1)
⟹ h = 50√3(√3 + 1)/(3 − 1)
⟹ h = 50√3(√3 + 1)/2
⟹ h = 50(3 + √3)/2
⟹ h = 25(3 + √3) m
⟹ h = (75 + 25√3) m

Alternatively: h = 50(√3 + 1)/(√3 − 1) · (√3 + 1)/(√3 + 1) = 50(√3 + 1)²/(3 − 1) = 25(√3 + 1)² = 25(3 + 2√3 + 1) = 25(4 + 2√3) = 50(2 + √3) m

Simplified: h = 50(√3 + 1)/(√3 − 1) = 50(√3 + 1)²/2 = 25(4 + 2√3) = 50 + 25√3 m

Question 7

A man standing at a point P on level ground observes the angle of elevation to the top of a vertical pole to be 30°. He walks 20 m directly towards the base of the pole and observes the angle of elevation to be 60°. Find the height of the pole.

Accepted Answer

Setup: This is a classic two-position heights and distances problem using the tangent ratio in right triangles.

Let h = height of the pole, and let the initial distance from P to the base of the pole be d metres.

From position P:
 tan(30°) = h/d
 1/√3 = h/d
 d = h√3 ... (1)

After walking 20 m towards the pole, the distance becomes (d − 20):
 tan(60°) = h/(d − 20)
 √3 = h/(d − 20)
 d − 20 = h/√3 ... (2)

Substituting (1) into (2):
 h√3 − 20 = h/√3
 h√3 − h/√3 = 20
 h(√3 − 1/√3) = 20
 h(3/√3 − 1/√3) = 20
 h(2/√3) = 20
 h = 20√3/2 = 10√3 metres

Verification: d = 10√3 · √3 = 30 m. After walking 20 m: new distance = 10 m. tan(60°) = 10√3/10 = √3 ✓

Question 8

From the top of a cliff 100 m high, the angle of depression to a boat on the water is 30°. How far is the boat from the base of the cliff?

Accepted Answer

Setup: Angle of depression problem — the angle measured downward from the horizontal at the observer's position equals the angle of elevation from the boat to the cliff top (alternate interior angles).

Let d = horizontal distance from the base of the cliff to the boat.
Height of cliff = 100 m.
Angle of depression = 30°.

From the top of the cliff, the angle of depression of 30° means the angle of elevation from the boat is also 30°.

Using the tangent ratio in the right triangle:
 tan(30°) = opposite/adjacent = 100/d
 1/√3 = 100/d
 d = 100√3 metres

Alternatively, using angle of depression directly from the cliff top:
 tan(30°) = 100/d
 d = 100/tan(30°) = 100/(1/√3) = 100√3 metres ≈ 173.2 m

Question 9

Two buildings of heights 20 m and 30 m stand on level ground 50 m apart. Find the angle of elevation from the top of the shorter building to the top of the taller building.

Accepted Answer

Setup: This problem requires setting up a right triangle using the difference in heights and the horizontal distance between buildings.

Height of shorter building = 20 m.
Height of taller building = 30 m.
Horizontal distance between buildings = 50 m.

The vertical difference in heights = 30 − 20 = 10 m.

From the top of the shorter building, we look up to the top of the taller building. The line of sight forms a right triangle where:
- The vertical leg (opposite to the angle of elevation) = 10 m
- The horizontal leg (adjacent to the angle of elevation) = 50 m

Let θ = angle of elevation.
 tan(θ) = opposite/adjacent = 10/50 = 1/5
 θ = arctan(1/5) ≈ 11.31°

Alternatively, we can express this as:
 tan(θ) = 1/5, so θ = tan⁻¹(1/5)

The exact answer is θ = tan⁻¹(1/5) or approximately 11.31°.

Question 10

A ladder 13 m long leans against a vertical wall. If the angle between the ladder and the ground is 60°, how high up the wall does the ladder reach?

Accepted Answer

Setup: Right triangle problem with the ladder as the hypotenuse, the wall as the vertical leg, and the ground as the horizontal leg.

Length of ladder (hypotenuse) = 13 m.
Angle between ladder and ground = 60°.
Height reached on wall = h (opposite to the 60° angle).

Using the sine ratio:
 sin(60°) = opposite/hypotenuse = h/13
 √3/2 = h/13
 h = 13√3/2 metres

Alternatively, using the cosine ratio for the base:
 cos(60°) = adjacent/hypotenuse = base/13
 1/2 = base/13
 base = 13/2 = 6.5 m

Then using the Pythagorean theorem:
 h² + (6.5)² = 13²
 h² + 42.25 = 169
 h² = 126.75 = 507/4
 h = √(507/4) = (√507)/2 = (13√3)/2 metres ≈ 11.26 m

The height is 13√3/2 metres or 6.5√3 metres.

Question 11

A ladder leans against a vertical wall. The angle of elevation from a point on the ground 3 m from the wall to the top of the ladder is 60°. If the ladder makes an angle of 45° with the ground, find the length of the ladder.

Accepted Answer

Setup: This problem combines angle of elevation with the angle the ladder makes with the ground.

Let L = length of the ladder.
The ladder makes an angle of 45° with the ground.

When the ladder leans against the wall at angle 45° with the ground:
  Height reached on wall: h = L·sin(45°) = L/√2
  Distance from wall to base of ladder: d = L·cos(45°) = L/√2

However, we are told that from a point 3 m from the wall, the angle of elevation to the top of the ladder is 60°.

This means the observer is 3 m from the wall (not at the base of the ladder).

From the observer's position (3 m from wall):
  tan(60°) = h/3
  √3 = h/3
  h = 3√3 m

Since the ladder reaches height h = 3√3 m and makes angle 45° with ground:
  h = L·sin(45°)
  3√3 = L · (1/√2)
  L = 3√3 · √2 = 3√6 m

Verification:
  Distance from wall to ladder base: d = L·cos(45°) = 3√6 · (1/√2) = 3√6/√2 = 3√3 m
  Height: h = 3√3 m
  From 3 m away: tan(angle) = 3√3/3 = √3, so angle = 60° ✓

Question 12

From the top of a building, the angles of depression to the top and bottom of a vertical pole standing on the same level ground are 30° and 60° respectively. If the building is 20 m tall, find the height of the pole.

Accepted Answer

Setup: Angle of depression problem with two observations to different parts of a vertical pole.

Let:
  H = 20 m (height of building)
  h = height of pole (to be found)
  d = horizontal distance from building to pole

Angle of depression to the bottom of the pole is 60°:
  tan(60°) = H/d
  √3 = 20/d
  d = 20/√3 = 20√3/3 m  ... (1)

Angle of depression to the top of the pole is 30°:
  The vertical distance from top of building to top of pole is (H − h) = (20 − h)
  tan(30°) = (H − h)/d
  1/√3 = (20 − h)/d
  d = √3(20 − h)  ... (2)

Equating (1) and (2):
  20√3/3 = √3(20 − h)
  20/3 = 20 − h
  h = 20 − 20/3
  h = (60 − 20)/3
  h = 40/3 m

Verification:
  d = 20√3/3 m
  tan(60°) = 20/(20√3/3) = 20 · 3/(20√3) = 3/√3 = √3 ✓
  tan(30°) = (20 − 40/3)/(20√3/3) = (60/3 − 40/3)/(20√3/3) = (20/3)/(20√3/3) = 1/√3 ✓

Question 13

From the top of a cliff of height 100 m, the angles of depression to two boats on the water are 30° and 60° respectively. If the boats are on the same side of the cliff and in line with the foot of the cliff, find the distance between the boats.

Accepted Answer

Setup: Angle of depression from a point equals the angle of elevation from the observer's perspective. We use the tangent ratio to find horizontal distances from the cliff base to each boat.

Let the cliff height be H = 100 m. Let the distances from the foot of the cliff to the two boats be d₁ and d₂, where d₁ > d₂ (boat 1 is farther).

For the boat with angle of depression 30°:
  tan(30°) = H/d₁
  1/√3 = 100/d₁
  d₁ = 100√3 m

For the boat with angle of depression 60°:
  tan(60°) = H/d₂
  √3 = 100/d₂
  d₂ = 100/√3 = 100√3/3 m

Distance between boats:
  Δd = d₁ − d₂ = 100√3 − 100√3/3
  Δd = 100√3(1 − 1/3)
  Δd = 100√3 · 2/3
  Δd = 200√3/3 m

Alternatively: Δd = 100√3 − 100√3/3 = (300√3 − 100√3)/3 = 200√3/3 m ≈ 115.47 m

Question 14

A man standing at a point P observes the angle of elevation to the top of a vertical tower AB to be 30°. He then walks 40 m towards the base of the tower along level ground and observes the angle of elevation to be 60°. Find the height of the tower.

Accepted Answer

Setup: This is a two-position angle of elevation problem requiring the use of tangent ratios and algebraic elimination.

Let h = height of tower AB, and let the initial distance from P to the base B be d metres.

From position P:
  tan(30°) = h/d
  1/√3 = h/d
  d = h√3  ... (1)

After walking 40 m towards B, the new distance is (d − 40):
  tan(60°) = h/(d − 40)
  √3 = h/(d − 40)
  d − 40 = h/√3  ... (2)

Substituting (1) into (2):
  h√3 − 40 = h/√3
  h√3 − h/√3 = 40
  h(√3 − 1/√3) = 40
  h(3/√3 − 1/√3) = 40
  h(2/√3) = 40
  h = 40√3/2 = 20√3 m

Verification: d = 20√3 · √3 = 60 m. Check: tan(30°) = 20√3/60 = √3/3 = 1/√3 ✓
After walking 40 m: distance = 20 m. tan(60°) = 20√3/20 = √3 ✓

Question 15

A vertical pole of height 12 m stands on level ground. A man at a distance x metres from the base of the pole observes the angle of elevation to the top to be 45°. Another man at a distance y metres from the base observes the angle of elevation to be 30°. If x and y are both positive and x < y, find the value of (y − x).

Accepted Answer

Setup: Two observers at different distances from a vertical pole observe different angles of elevation. We use the tangent ratio to relate distances to the known height.

Let the pole height be h = 12 m.

For the first man at distance x:
  tan(45°) = h/x
  1 = 12/x
  x = 12 m

For the second man at distance y:
  tan(30°) = h/y
  1/√3 = 12/y
  y = 12√3 m

Verify x < y: 12 < 12√3 ✓ (since √3 ≈ 1.732)

Therefore:
  y − x = 12√3 − 12
  y − x = 12(√3 − 1) m

Numerically: y − x ≈ 12(1.732 − 1) = 12(0.732) ≈ 8.78 m

Question 16

From the top of a cliff 100 m high, the angles of depression to two boats on the water are 30° and 60° respectively. If the boats are on the same side of the cliff and in line with the foot of the cliff, find the distance between the boats.

Accepted Answer

Setup: Angle of depression problem with two objects at different distances from the base of a vertical cliff.

Let the cliff height be h = 100 m. Let the two boats be at distances d₁ and d₂ from the foot of the cliff (with d₁ < d₂, since the angle of depression to the nearer boat is larger).

Angle of depression equals angle of elevation from the boat's perspective.

For the boat with angle of depression 60°:
  tan(60°) = h/d₁
  √3 = 100/d₁
  d₁ = 100/√3 = 100√3/3 m  ... (1)

For the boat with angle of depression 30°:
  tan(30°) = h/d₂
  1/√3 = 100/d₂
  d₂ = 100√3 m  ... (2)

Distance between boats:
  d₂ − d₁ = 100√3 − 100√3/3
  = 100√3(1 − 1/3)
  = 100√3 · (2/3)
  = 200√3/3 m

Alternatively: d₂ − d₁ = (300√3 − 100√3)/3 = 200√3/3 m ≈ 115.47 m

Question 17

A person standing at point A on level ground observes the angle of elevation to the top of a vertical tower at point B as 30°. After walking 40 m towards the base of the tower (to point C), the angle of elevation becomes 60°. Find the height of the tower.

Accepted Answer

Setup: This is a two-position angle of elevation problem requiring application of trigonometric ratios in right triangles.

Let h = height of the tower, and let the initial distance from A to the base of the tower be d.

From position A:
  tan(30°) = h/d
  1/√3 = h/d
  d = h√3 ... (1)

From position C (after walking 40 m closer):
  tan(60°) = h/(d - 40)
  √3 = h/(d - 40)
  d - 40 = h/√3 ... (2)

Substituting (1) into (2):
  h√3 - 40 = h/√3
  h√3 - h/√3 = 40
  h(√3 - 1/√3) = 40
  h(3/√3 - 1/√3) = 40
  h(2/√3) = 40
  h = 40√3/2 = 20√3 m

Verification: d = 20√3 · √3 = 60 m. From C: tan(60°) = 20√3/20 = √3 ✓

Question 18

From the top of a lighthouse 60 m high, the angles of depression of two ships on the same side of the lighthouse are 60° and 30°. What is the distance between the two ships?

Accepted Answer

Let the lighthouse AB = 60 m (A at top, B at base). Let C and D be the positions of the two ships on the same side, with angles of depression 60° and 30° respectively.

Step 1: Find BC (nearer ship).
tan 60° = AB/BC → √3 = 60/BC → BC = 60/√3 = 20√3 m.

Step 2: Find BD (farther ship).
tan 30° = AB/BD → 1/√3 = 60/BD → BD = 60√3 m.

Step 3: Distance between ships = BD − BC = 60√3 − 20√3 = 40√3 m.

Rule used: In right triangle, tan(angle of depression) = height / horizontal distance.

Question 19

From the top of a cliff of height 100 m, the angle of depression to a boat at sea is 30°. A second boat is directly behind the first, and the angle of depression to the second boat is 60°. Find the distance between the two boats (assume they are on the same horizontal line).

Accepted Answer

Setup: Angle of depression problem with two objects at different horizontal distances from the base of the cliff.

Let the cliff height be H = 100 m. Let d₁ = horizontal distance from cliff base to first boat, and d₂ = horizontal distance from cliff base to second boat.

Angle of depression equals angle of elevation from the boat's perspective.

For the first boat (angle of depression 30°):
  tan(30°) = H/d₁
  1/√3 = 100/d₁
  d₁ = 100√3 m ... (1)

For the second boat (angle of depression 60°):
  tan(60°) = H/d₂
  √3 = 100/d₂
  d₂ = 100/√3 = 100√3/3 m ... (2)

Distance between boats:
  Δd = d₁ − d₂ = 100√3 − 100√3/3
  Δd = 100√3(1 − 1/3)
  Δd = 100√3 · 2/3
  Δd = 200√3/3 m

Numerical check: 200√3/3 ≈ 200(1.732)/3 ≈ 115.5 m

Question 20

A vertical pole of height 12 m stands on level ground. An observer at point P on the ground sees the top of the pole at an angle of elevation α, where tan(α) = 3/4. The observer then moves 5 m further away from the pole (to point Q), and the angle of elevation becomes β. If tan(β) = 1/2, verify which of the following is the distance PQ.

Accepted Answer

Setup: Two-position problem where we verify consistency using the tangent values given.

Let h = 12 m (pole height), and let x = distance from P to the base of the pole.

At position P:
  tan(α) = h/x = 12/x
  3/4 = 12/x
  x = 16 m ... (1)

At position Q (5 m further away):
  tan(β) = h/(x + 5) = 12/(x + 5)
  1/2 = 12/(x + 5)
  x + 5 = 24
  x = 19 m ... (2)

Inconsistency check: From (1), x = 16. From (2), x = 19. These do not match.

However, the question asks for distance PQ. If we assume the given angles are correct:
  From P: distance to pole = 16 m
  From Q: distance to pole = 24 m
  Distance PQ = 24 − 16 = 8 m

Note: The problem statement contains a physical inconsistency (the observer cannot move 5 m and have both angle conditions satisfied simultaneously with these exact values). However, the mathematical answer for the distance between the two positions is 8 m.

NDA Heights & Distances

Heights & Distances— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Heights & Distances — Practice Questions

60-Second Revision — Heights & Distances