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Angles of Elevation and Depression

Classic word problems — towers, cliffs, lighthouses, kites, ladders — each with a clear diagram and tan or sin.

What the Angles Mean

These angles show up everywhere in textbooks: looking up at a tower, down from a lighthouse to a boat, flying a kite, or leaning a ladder on a wall. Both are measured from a horizontal line through the observer's eye — never from a wall or the sloping line of sight alone.

Two right triangles: elevation up from the horizontal base, depression down from a horizontal line at the top. — $Both angles are measured from the horizontal . Opposite = vertical, adjacent = horizontal.$

In almost every problem you get a right-angled triangle where: the vertical side is opposite the angle the horizontal side is adjacent to the angle So you usually use tan : tan (θ) = \frac{opposite}{adjacent} When the sloping length is given (string, ladder), use sin or cos instead.

Elevation — Looking Up

Start with height and distance problems. Draw the triangle, mark θ at the observer, then use tan.

Right triangle: adjacent 20 m, elevation 35°, opposite labelled h. — $Known adjacent + elevation \to find opposite with an .$

Example 1 — height of a tower

A person stands 20 m from the base of a tower. The angle of elevation to the top is 35 °. Find the height of the tower. Step 1 . Adjacent = 20 m, opposite = height, θ = 35 °. Step 2 . tan (3 5^{\circ}) = \frac{height}{20} Step 3 . tan (3 5^{\circ}) \approx 0.7002, so height = 20 \times 0.7002 \approx 14.0 m . Answer: \approx 14.0 m

Right triangle: adjacent 80 m, elevation 38°, opposite labelled h. — $Same method: an (h e t a) = df r a c h e x t a d j a ce n t .$

Example 1 — height of a cliff

From a point 80 m from the base of a cliff, the angle of elevation to the top is 38 °. How high is the cliff? Step 1 . tan (3 8^{\circ}) = \frac{h}{80} Step 2 . tan (3 8^{\circ}) \approx 0.7813, so h = 80 \times 0.7813 \approx 62.5 m . Answer: \approx 62.5 m (about 63 m)

Right triangle: 40 m string as hypotenuse, 50° between the ground and the string, unknown height h. — $Elevation = angle between the ground and the string. String = hypotenuse \to use sin .$

Example 1 — kite on a string

A kite string is 40 m long and makes a 50 ° angle of elevation with the ground. How high is the kite? Step 1 . Hypotenuse = 40 m, opposite = height, θ = 50 °. Step 2 . sin (5 0^{\circ}) = \frac{h}{40} Step 3 . sin (5 0^{\circ}) \approx 0.7660, so h = 40 \times 0.7660 \approx 30.6 m . Answer: \approx 30.6 m

Right triangle: hypotenuse 3.5 m, elevation 48°, opposite labelled h. — $Same as Example 3 : s in (h e t a) = df r a c h e x t h y p o t e n u se .$

Example 1 — ladder against a wall

A 3.5 m ladder leans against a wall and makes a 48 ° angle with the ground. How high up the wall does it reach? Step 1 . Hypotenuse = 3.5 m, opposite = height on wall, θ = 48 °. Step 2 . sin (4 8^{\circ}) = \frac{h}{3.5} Step 3 . sin (4 8^{\circ}) \approx 0.7431, so h = 3.5 \times 0.7431 \approx 2.60 m . Answer: \approx 2.60 m

Angle of elevation

Adjacent 30 mElevation 50°

a ≈ 35.75 m · total = a + 1.7 ≈ 37.45 m

Change a value and watch:

Height above eyes = distance \times tan(angle). Total height = that plus eye height.

Interactive lab — drag distance and elevation angle.

Depression — Looking Down

From a high place, draw the horizontal at eye level, then the line of sight down. Depression from the top equals elevation from the object (alternate angles with parallel horizontals).

Right triangle: opposite 40 m, depression 28° from horizontal, adjacent labelled d. — $Depression equals elevation at the base. Use an .$

Example 1 — boat from a lighthouse

From the top of a 40 m lighthouse, the angle of depression to a boat is 28 °. How far is the boat from the base of the lighthouse? Step 1 . Depression 28 ° = elevation 28 ° from the boat. Step 2 . tan (2 8^{\circ}) = \frac{40}{d} Step 3 . tan (2 8^{\circ}) \approx 0.5317, so d = \frac{40}{0.5317} \approx 75.2 m . Answer: \approx 75.2 m

Right triangle: opposite 50 m, depression 25° (and matching 25° at the base), adjacent labelled d. — $Same setup: an (h e t a) = df r a c e x t h e i g h t d .$

Example 1 — boat from a cliff

From the top of a 50 m cliff, the angle of depression to a boat is 25 °. How far is the boat from the base of the cliff? Step 1 . tan (2 5^{\circ}) = \frac{50}{d} Step 2 . tan (2 5^{\circ}) \approx 0.4663, so d = \frac{50}{0.4663} \approx 107.2 m . Answer: \approx 107.2 m

Bridge 18 m above water, depression 30° to a buoy; distance about 31.2 m. — $Special angle 30 ° lets you leave the answer in exact form with 3 .$

Example 1 — buoy from a bridge

From a bridge 18 m above a river, the angle of depression to a buoy is 30 °. How far is the buoy from the point on the water directly below the observer? (Use tan 3 0^{\circ} = \frac{1}{3} .) Step 1 . tan (3 0^{\circ}) = \frac{18}{d} Step 2 . d = \frac{18}{tan ( 3 0 ^{\circ} )} = 183 \approx 31.2 m Answer: \approx 31.2 m (exact: 183 m)

Angle of depression

Opposite (height) 40 mDepression 28°

Adjacent d ≈ 75.23 m

Change a value and watch:

Horizontal distance = \frac{height}{tan ( depression angle )} .

Interactive lab — steeper depression means the boat is closer.

When Eye Height Matters

The triangle starts at eye level, not at the ground. If the question asks for the full height of the object, add eye height after you find the height above the eyes.

Right triangle above eye level: adjacent 30 m, elevation 50°, opposite a, plus 1.7 m eye height. — $Find height above eyes first, then add eye height for the total.$

Example 1 — flagpole with eye height

From 30 m away, the angle of elevation to a flagpole top is 50 °. Eyes are 1.7 m above the ground. How tall is the flagpole? (tan 5 0^{\circ} \approx 1.1918) Step 1 — height above eyes only. above eyes = 30 \times tan (5 0^{\circ}) \approx 30 \times 1.1918 \approx 35.75 m Step 2 — add eye height. Total = 35.75 + 1.7 \approx 37.5 m Answer: \approx 37.5 m

Watch Outs

Practice

Swipe or use the arrows to move between questions.

1 / 3

From 25 m away, the angle of elevation to a tower top is 35 °. Eyes are 1.7 m above ground. Tower height to 1 d. p .? (tan 35 ° \approx 0.700)

Your answer

From a 36 m cliff, depression to a boat is 40 °. Distance from the cliff base to the boat, to 1 d. p .? (tan 40 ° \approx 0.839)

Your answer

A kite string is 25 m long at 40 ° elevation. How high is the kite, to 1 d. p .? (sin 40 ° \approx 0.6428)

Your answer

More lessons in Trigonometry · Next: Trigonometry Foundations Worksheet · Previous: Finding Missing Angles

What the Angles Mean

These angles show up everywhere in textbooks: looking up at a tower, down from a lighthouse to a boat, flying a kite, or leaning a ladder on a wall. Both are measured from a horizontal line through the observer's eye — never from a wall or the sloping line of sight alone.

In almost every problem you get a right-angled triangle where: the vertical side is opposite the angle the horizontal side is adjacent to the angle So you usually use tan : tan (θ) = \frac{opposite}{adjacent} When the sloping length is given (string, ladder), use sin or cos instead.

Elevation — Looking Up

Start with height and distance problems. Draw the triangle, mark θ at the observer, then use tan.

Example 1 — height of a tower

A person stands 20 m from the base of a tower. The angle of elevation to the top is 35 °. Find the height of the tower. Step 1 . Adjacent = 20 m, opposite = height, θ = 35 °. Step 2 . tan (3 5^{\circ}) = \frac{height}{20} Step 3 . tan (3 5^{\circ}) \approx 0.7002, so height = 20 \times 0.7002 \approx 14.0 m . Answer: \approx 14.0 m

Example 1 — height of a cliff

From a point 80 m from the base of a cliff, the angle of elevation to the top is 38 °. How high is the cliff? Step 1 . tan (3 8^{\circ}) = \frac{h}{80} Step 2 . tan (3 8^{\circ}) \approx 0.7813, so h = 80 \times 0.7813 \approx 62.5 m . Answer: \approx 62.5 m (about 63 m)

Example 1 — kite on a string

A kite string is 40 m long and makes a 50 ° angle of elevation with the ground. How high is the kite? Step 1 . Hypotenuse = 40 m, opposite = height, θ = 50 °. Step 2 . sin (5 0^{\circ}) = \frac{h}{40} Step 3 . sin (5 0^{\circ}) \approx 0.7660, so h = 40 \times 0.7660 \approx 30.6 m . Answer: \approx 30.6 m

Example 1 — ladder against a wall

A 3.5 m ladder leans against a wall and makes a 48 ° angle with the ground. How high up the wall does it reach? Step 1 . Hypotenuse = 3.5 m, opposite = height on wall, θ = 48 °. Step 2 . sin (4 8^{\circ}) = \frac{h}{3.5} Step 3 . sin (4 8^{\circ}) \approx 0.7431, so h = 3.5 \times 0.7431 \approx 2.60 m . Answer: \approx 2.60 m

Angle of elevation

Adjacent 30 mElevation 50°

a ≈ 35.75 m · total = a + 1.7 ≈ 37.45 m

Change a value and watch:

Height above eyes = distance \times tan(angle). Total height = that plus eye height.

Interactive lab — drag distance and elevation angle.

Depression — Looking Down

From a high place, draw the horizontal at eye level, then the line of sight down. Depression from the top equals elevation from the object (alternate angles with parallel horizontals).

Example 1 — boat from a lighthouse

From the top of a 40 m lighthouse, the angle of depression to a boat is 28 °. How far is the boat from the base of the lighthouse? Step 1 . Depression 28 ° = elevation 28 ° from the boat. Step 2 . tan (2 8^{\circ}) = \frac{40}{d} Step 3 . tan (2 8^{\circ}) \approx 0.5317, so d = \frac{40}{0.5317} \approx 75.2 m . Answer: \approx 75.2 m

Example 1 — boat from a cliff

From the top of a 50 m cliff, the angle of depression to a boat is 25 °. How far is the boat from the base of the cliff? Step 1 . tan (2 5^{\circ}) = \frac{50}{d} Step 2 . tan (2 5^{\circ}) \approx 0.4663, so d = \frac{50}{0.4663} \approx 107.2 m . Answer: \approx 107.2 m

Example 1 — buoy from a bridge

From a bridge 18 m above a river, the angle of depression to a buoy is 30 °. How far is the buoy from the point on the water directly below the observer? (Use tan 3 0^{\circ} = \frac{1}{3} .) Step 1 . tan (3 0^{\circ}) = \frac{18}{d} Step 2 . d = \frac{18}{tan ( 3 0 ^{\circ} )} = 183 \approx 31.2 m Answer: \approx 31.2 m (exact: 183 m)

Angle of depression

Opposite (height) 40 mDepression 28°

Adjacent d ≈ 75.23 m

Change a value and watch:

Horizontal distance = \frac{height}{tan ( depression angle )} .

Interactive lab — steeper depression means the boat is closer.

When Eye Height Matters

The triangle starts at eye level, not at the ground. If the question asks for the full height of the object, add eye height after you find the height above the eyes.

Example 1 — flagpole with eye height

From 30 m away, the angle of elevation to a flagpole top is 50 °. Eyes are 1.7 m above the ground. How tall is the flagpole? (tan 5 0^{\circ} \approx 1.1918) Step 1 — height above eyes only. above eyes = 30 \times tan (5 0^{\circ}) \approx 30 \times 1.1918 \approx 35.75 m Step 2 — add eye height. Total = 35.75 + 1.7 \approx 37.5 m Answer: \approx 37.5 m

Watch Outs

Practice

Swipe or use the arrows to move between questions.

1 / 3

From 25 m away, the angle of elevation to a tower top is 35 °. Eyes are 1.7 m above ground. Tower height to 1 d. p .? (tan 35 ° \approx 0.700)

Your answer

From a 36 m cliff, depression to a boat is 40 °. Distance from the cliff base to the boat, to 1 d. p .? (tan 40 ° \approx 0.839)

Your answer

A kite string is 25 m long at 40 ° elevation. How high is the kite, to 1 d. p .? (sin 40 ° \approx 0.6428)

Your answer