MathsCBSE Class 10 · Some Applications of Trigonometry

Angle of Depression Class 10: Meaning, Proof and Solved Questions

Updated 14 October 2026 · Fact-checked

The angle of depression is the angle between the horizontal line through an observer's eye and the line of sight to an object below eye level. It equals the angle of elevation of the observer seen from the object, by alternate angles. The board asks you to transfer this angle correctly and find heights or distances.

Understand Angle of Depression

When you stand on a cliff, a terrace or a lighthouse and look down at a boat or a car, you lower your head. Imagine a horizontal line straight out from your eye. The angle between that horizontal line and your line of sight going down to the object is the angle of depression.

It is always measured from the horizontal at the observer, never from the vertical side of the building. A common slip is to mark it inside the triangle at the top, between the wall and the line of sight. That angle is actually $9 0^{\circ}$ minus the depression.

Why the angles are equal: the horizontal through your eye is parallel to the ground. The line of sight is a transversal cutting both lines. So the angle of depression at the top equals the angle the line of sight makes with the ground at the object. These are alternate interior angles. In words: the angle of depression of B seen from A equals the angle of elevation of A seen from B.

Why it is useful: this turns every depression problem into an elevation problem. Mark the depression at the top, transfer it to the ground point with the reason "alternate angles", and you have an ordinary right triangle with the right angle at the foot of the building. Then use $tan$ for height and ground distance, or $sin$ and $cos$ when the slanted line of sight is involved.

The figure in words: draw the building as a vertical line AB, top A and foot B on the ground line. Draw a dotted horizontal line from A, parallel to the ground. Join A to the object C on the ground; the angle between the dotted line and AC is the depression, and the same angle sits at C between the ground and CA. For two objects, put both on the ground line: on the same side of B you subtract their distances from B, and on opposite sides you add them.

The result holds only when the ground is level (horizontal), so that it is parallel to the observer's horizontal line. For the board, the angles will be $3 0^{\circ}$ , $4 5^{\circ}$ or $6 0^{\circ}$ and at most two right triangles. A larger angle of depression means the object is closer to the foot of the building, because you must look down more steeply.

Angle of Depression: formulas to remember

Depression equals elevation: $angle of depression of B from A = angle of elevation of A from B$; Reason: horizontal at the eye is parallel to level ground, so alternate angles are equal. Write this reason.
Distance from the foot: $d = \frac{h}{t a n θ} = h cot θ$; h is the height of the observer's eye above the ground, θ the angle of depression.
Line of sight: $l = \frac{h}{s i n θ}$; Slanted distance from observer to object.
Angle inside the triangle at the top: $∠ (vertical, line of sight) = 9 0^{\circ} - θ$; Not the angle of depression. Avoid using it by mistake.
Distances from the line of sight: $h = l sin θ, d = l cos θ$; l is the line of sight, θ the angle of depression, h the height and d the ground distance.
Two objects, same side: $C D = h (cot θ_{1} - cot θ_{2})$; θ₁ is the smaller angle of depression (farther object), θ₂ the larger (nearer object).
Two objects, opposite sides: $C D = h (cot θ_{1} + cot θ_{2})$; Each object is measured from the foot, so the distances add.

How to solve Angle of Depression questions

Use this routine for any depression question, with one object or two.

1Draw the ground as a horizontal line and the building, cliff or tower as a vertical line; label top A and foot B.
2Draw a dotted horizontal line through A, parallel to the ground.
3Mark each angle of depression between this horizontal and the line of sight to the object.
4Transfer each angle to the object's position on the ground and write: alternate angles, horizontal is parallel to the ground.
5In each right triangle (right angle at B), pick tan for height and distance, sin or cos if the line of sight is wanted.
6For two objects, decide same side (subtract distances) or opposite sides (add distances).
7Rationalise, substitute the given value of root 3 only at the end, and write units.

Angle of Depression in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Angle of depression

Suppose you are on the top-floor balcony of a hostel above a level playground. You see two friends standing on a straight line from the foot of the hostel, on the same side, 32 m apart. The angles of depression of the two friends are $6 0^{\circ}$ and $3 0^{\circ}$ .

How high is your balcony, and how far is the nearer friend from the foot of the hostel? Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the balcony be A, the foot B, the near friend C and the far friend D; let AB = h m.
By alternate angles, $∠ A C B = 6 0^{\circ}$ and $∠ A D B = 3 0^{\circ}$ .
$B C = \frac{h}{t a n 6 0 ^{\circ}} = \frac{h}{3}$ and $B D = \frac{h}{t a n 3 0 ^{\circ}} = 3 h$ .
Same side, so $B D - B C = 32$ : $3 h - \frac{h}{3} = \frac{2 h}{3} = 32$ .
$h = 163 = 16 \times 1.732 = 27.712$ m.
Nearer friend: $B C = \frac{16 3}{3} = 16$ m. Check: $B D = 163 \times 3 = 48$ m and $48 - 16 = 32$ m.

Answer: The balcony is $163 \approx 27.71$ m high; the nearer friend is 16 m from the foot.

Memory hook: Looking down from the hostel: bigger depression, closer friend.

Angle of depression

Imagine a water tank on a level terrace, its top 20 m above the road. A rickshaw on the road moves straight towards the building. From the tank top, its angle of depression changes from $4 5^{\circ}$ to $6 0^{\circ}$ in 4 seconds.

Find the distance the rickshaw moved, its speed in m/s and in km/h, and how much longer it takes to reach the foot of the building at the same speed. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

First distance from foot: $\frac{20}{t a n 4 5 ^{\circ}} = 20$ m.
Later distance: $\frac{20}{t a n 6 0 ^{\circ}} = \frac{20}{3} = \frac{20 3}{3} \approx 11.547$ m.
Distance moved $= 20 - 11.547 = 8.453$ m.
Speed $= \frac{8.453}{4} \approx 2.11$ m/s, which is about $2.11 \times 3.6 \approx 7.6$ km/h.
Further time $= \frac{11.547}{2.113} \approx 5.46$ s. Exactly: $\frac{20/ 3}{20 ( 1 - 1/ 3 ) /4} = \frac{4}{3 - 1} = 2 (3 + 1) \approx 5.46$ s.

Answer: It moved about 8.45 m at about 2.11 m/s (7.6 km/h), and needs about 5.46 s more to reach the foot.

Memory hook: Tank top watches the rickshaw: angle up, gap down.

Tricks to remember Angle of Depression

Short memory aids. Say each one aloud once, then use it on a question the same day.

Depression is drawn on the horizontal
Eyes straight ahead first, then look down: the angle starts from where your eyes were, the horizontal.
Use it when: Stops you marking the angle against the wall.
Z for transfer
The horizontal, the line of sight and the ground make a Z shape; the angles in the corners of a Z are equal.
Use it when: Reminds you of alternate angles and the reason to write.
Steep means near
Big depression, look nearly straight down, so the object is close to the foot.
Use it when: Quick check that your distances are in the right order.

Angle of Depression: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Boat from a lighthouse · 2 marks · Very short answerBoard level

From the top of a lighthouse 50 m high on level ground at the sea shore, the angle of depression of a ship is $3 0^{\circ}$ . Find the distance of the ship from the foot of the lighthouse and the length of the line of sight. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let AB = 50 m be the lighthouse, C the ship. By alternate angles, $∠ A C B = 3 0^{\circ}$ . (½)
$tan 3 0^{\circ} = \frac{A B}{B C}$ , so $B C = 503 = 86.6$ m. (½)
$sin 3 0^{\circ} = \frac{A B}{A C}$ , so $A C = \frac{50}{1/2} = 100$ m. (1)

Answer: Distance $503 \approx 86.6$ m; line of sight 100 m.

Example 2: Two cars on the same side · 3 marks · Short answerBoard level

From the top of a tower 75 m high, the angles of depression of two cars on the same side of the tower, in line with its foot, are $3 0^{\circ}$ and $4 5^{\circ}$ . Find the distance between the cars. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Tower AB = 75 m; near car C, far car D. By alternate angles, $∠ A C B = 4 5^{\circ}$ , $∠ A D B = 3 0^{\circ}$ . (½)
$tan 4 5^{\circ} = \frac{75}{B C}$ , so $B C = 75$ m. (1)
$tan 3 0^{\circ} = \frac{75}{B D}$ , so $B D = 753$ m. (½)
$C D = 753 - 75 = 75 (3 - 1) = 75 \times 0.732 = 54.9$ m. (1)

Answer: $75 (3 - 1) \approx 54.9$ m

Example 3: Depressions of top and foot of a pole · 3 marks · Short answerBoard level

From the top of a building 60 m high, the angles of depression of the top and the foot of a vertical pole on the same level ground are $3 0^{\circ}$ and $6 0^{\circ}$ . Find the height of the pole and the distance between the building and the pole.

Try it first, then show the solutionSolution

Building AB = 60 m, pole CD with foot D, top C. By alternate angles, $∠ A D B = 6 0^{\circ}$ . (½)
$B D = \frac{60}{3} = 203$ m. (1)
Draw CE horizontal to AB, so $C E = 203$ m and $∠ A C E = 3 0^{\circ}$ . $A E = 203 \times \frac{1}{3} = 20$ m. (1)
Pole $C D = E B = 60 - 20 = 40$ m. (½)

Answer: Pole 40 m; distance $203$ m (about 34.64 m)

Example 4: Car approaching a tower: time · 5 marks · Long answerHarder board level

From the top of a tower, the angle of depression of a car moving straight towards the tower at uniform speed changes from $3 0^{\circ}$ to $4 5^{\circ}$ in 12 seconds. Find the further time the car takes to reach the foot of the tower. Use $3 = 1.732$ . Then check whether the answer depends on the height of the tower.

Try it first, then show the solutionSolution

Let height AB = h. By alternate angles, the elevations of A from the car positions D and C are $3 0^{\circ}$ and $4 5^{\circ}$ . (1)
From C: $B C = \frac{h}{t a n 4 5 ^{\circ}} = h$ . (1)
From D: $B D = \frac{h}{t a n 3 0 ^{\circ}} = 3 h$ . (1)
$C D = (3 - 1) h$ covered in 12 s, so speed $= \frac{( 3 - 1 ) h}{12}$ . (1)
Time for BC $= \frac{h \times 12}{( 3 - 1 ) h} = \frac{12}{3 - 1} = 6 (3 + 1) = 6 \times 2.732 = 16.39$ s; h cancels, so the time does not depend on the height. (1)

Answer: $6 (3 + 1) \approx 16.39$ seconds; independent of the tower's height.

Example 5: Case study: coastguard tower · 4 marks · Case-basedHOTS / competency

A coastguard watches from a tower whose top is 40 m above sea level. A boat P is seen at an angle of depression of $6 0^{\circ}$ , and a second boat Q, directly behind P in the same line with the foot of the tower, at $3 0^{\circ}$ .
(i) What is the angle of elevation of the tower top seen from boat Q? Give the reason. (1)
(ii) Find the distance of boat P from the foot. (1)
(iii) Boat Q sails straight towards the tower at a uniform speed and reaches the position of P after 2 minutes. Find its speed in m/min and how much longer it will take to reach the foot of the tower. (2)

Try it first, then show the solutionSolution

(i) $3 0^{\circ}$ , since the horizontal at the top is parallel to the sea, so alternate angles are equal. (1)
(ii) $B P = \frac{40}{t a n 6 0 ^{\circ}} = \frac{40}{3} = \frac{40 3}{3} \approx 23.09$ m. (1)
(iii) $B Q = \frac{40}{t a n 3 0 ^{\circ}} = 403$ m, so $P Q = 403 - \frac{40 3}{3} = \frac{80 3}{3} \approx 46.19$ m. Speed $= \frac{80 3}{3} \div 2 = \frac{40 3}{3} \approx 23.09$ m/min. (1)
Remaining distance $B P = \frac{40 3}{3}$ m, which equals one minute of travel. So it takes 1 more minute. (1)

Answer: (i) $3 0^{\circ}$ (alternate angles); (ii) $\frac{40 3}{3} \approx 23.09$ m; (iii) speed $\frac{40 3}{3} \approx 23.09$ m/min, and 1 more minute to reach the foot.

Example 6: Two cars on opposite sides of a tower · 3 marks · Short answerBoard level

A tower AB stands on level ground, with top A and foot B, and AB = 60 m. Two cars C and D are on a straight road through B, on opposite sides of the tower. From A, the angle of depression of C is $6 0^{\circ}$ and that of D is $3 0^{\circ}$ . Find the distance between the two cars. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Draw the dotted horizontal through A. By alternate angles (horizontal parallel to the ground), $∠ A C B = 6 0^{\circ}$ and $∠ A D B = 3 0^{\circ}$ ; both triangles have the right angle at B. (½)
In right $△ A B C$ : $tan 6 0^{\circ} = \frac{60}{B C}$ , so $B C = \frac{60}{3} = 203$ m. (1)
In right $△ A B D$ : $tan 3 0^{\circ} = \frac{60}{B D}$ , so $B D = 603$ m. (1)
C and D are on opposite sides of B, so $C D = B C + B D = 203 + 603 = 803 = 80 \times 1.732 = 138.56$ m. (½)

Answer: $803 \approx 138.56$ m

Example 7: Balloon above a road: opposite sides · 5 marks · Long answerHarder board level

A balloon P is at height h metres vertically above a point B of a straight level road. Two points C and D on the road lie on opposite sides of B, and CD = 200 m. From the balloon, the angle of depression of C is $6 0^{\circ}$ and that of D is $3 0^{\circ}$ . Show that $h (\frac{1}{3} + 3) = 200$ and hence find h. Also find BC and BD. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: PB vertical with PB = h, C and D on the road on either side of B. By alternate angles (horizontal at P parallel to the road), $∠ P C B = 6 0^{\circ}$ and $∠ P D B = 3 0^{\circ}$ . (1)
In right $△ P B C$ : $tan 6 0^{\circ} = \frac{h}{B C}$ , so $B C = \frac{h}{3}$ . (1)
In right $△ P B D$ : $tan 3 0^{\circ} = \frac{h}{B D}$ , so $B D = 3 h$ . (1)
Opposite sides, so $B C + B D = C D$ : $h (\frac{1}{3} + 3) = 200$ . (1)
$\frac{4 h}{3} = 200$ , so $h = 503 = 86.6$ m; then $B C = 50$ m and $B D = 150$ m (check: $50 + 150 = 200$ ). (1)

Answer: $h = 503 \approx 86.6$ m; BC = 50 m, BD = 150 m

Example 8: Building and a shorter tower · 5 marks · Long answerHarder board level

A building AB (top A, foot B) and a tower CD (top C, foot D) stand on the same level ground. From A, the angles of depression of the top C and the foot D of the tower are $3 0^{\circ}$ and $4 5^{\circ}$ . The tower is 20 m shorter than the building. Find both heights and the distance between them. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the building be $h$ m high, so the tower is $(h - 20)$ m, and let the distance between them be $x$ m. Draw CE perpendicular to AB with E on AB. Then $C E = B D = x$ and $E B = C D = h - 20$ , so $A E = 20$ m. By alternate angles, $∠ A D B = 4 5^{\circ}$ and $∠ A C E = 3 0^{\circ}$ . (1)
In right $△ A B D$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $x = h$ . (1)
In right $△ A E C$ : $tan 3 0^{\circ} = \frac{A E}{C E} = \frac{20}{x}$ , so $x = 203$ m. (1)
Hence $h = x = 203 = 20 \times 1.732 = 34.64$ m. (1)
Tower $= h - 20 = 20 (3 - 1) = 14.64$ m; distance $B D = 203 \approx 34.64$ m. (1)

Answer: Building $203 \approx 34.64$ m; tower $20 (3 - 1) \approx 14.64$ m; distance $203 \approx 34.64$ m

Example 9: MCQ: depression of 45 degrees · 1 mark · MCQBoard level

From a point 10 m above level ground, the angle of depression of a stone on the ground is $4 5^{\circ}$ . The horizontal distance of the stone from the foot of the point of observation is
(a) $\frac{10}{3}$ m
(b) 10 m
(c) $103$ m
(d) 20 m

Try it first, then show the solutionSolution

By alternate angles the elevation from the stone is $4 5^{\circ}$ , so distance $= \frac{10}{t a n 4 5 ^{\circ}} = 10$ m. (1)

Answer: (b) 10 m

Example 10: Assertion-reason: depression and elevation · 1 mark · Assertion-reasonBoard level

Assertion (A): From the top of a 15 m building on level ground, the angle of depression of a point P on the ground is $3 0^{\circ}$ , so the angle of elevation of the top of the building from P is $3 0^{\circ}$ .
Reason (R): The horizontal line through the observer's eye is parallel to the level ground, so the alternate angles made by the line of sight are equal.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is true, and it is exactly why the depression at the top equals the elevation at P, so A is true and R explains it. (1)

Answer: (a) Both A and R are true and R is the correct explanation of A.

Where students lose marks in Angle of Depression

The slips examiners see most often, and what to write instead.

Mistake: Marking the depression against the vertical wall
Depression is from the horizontal; the angle at the wall is its complement.
Wrong working
$∠ B A C = 3 0^{\circ}$ , so $B C = A B tan 3 0^{\circ}$
Right working
$∠ A C B = 3 0^{\circ}$ (alternate angles), so $B C = \frac{A B}{t a n 3 0 ^{\circ}}$
Do this: Always draw the dotted horizontal at the top first.
Usually costs: Usually all marks after the figure.
Mistake: Not writing the reason for transferring the angle
Examiners expect the justification for the step.
Wrong working
$∠ A C B = 4 5^{\circ}$
Right working
$∠ A C B = 4 5^{\circ}$ (alternate angles, horizontal parallel to ground)
Do this: Write the reason in brackets once per figure.
Usually costs: Can cost ½ mark.
Mistake: Adding distances for objects on the same side
Same side means one triangle sits inside the other.
Wrong working
$C D = 75 + 753$
Right working
$C D = 753 - 75$
Do this: Check the figure: same side subtract, opposite sides add.
Usually costs: 1 to 2 marks.
Mistake: Rounding root 3 too early
Early decimals spread error through later steps.
Wrong working
$\frac{12}{0.73} = 16.4$
Right working
$\frac{12}{3 - 1} = 6 (3 + 1) = 16.39$
Do this: Rationalise in surd form, substitute at the last step.
Usually costs: Accuracy mark.

Exam tips

Draw a labelled figure with a dotted horizontal at the observer; it carries marks in long answers.
State the alternate angles reason the first time you move an angle to the ground.
For two objects, write which triangle each equation comes from, e.g. 'In right triangle ABC'.
Leave answers in surd form unless a value of root 3 is given.
In assertion-reason items, remember depression of B from A equals elevation of A from B only for level ground.

Practice questions on Angle of Depression

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 5 marks · Long answerHarder board level
A girl at the top of a tower 30 m high sees a car moving straight towards the foot of the tower at a uniform speed. The angle of depression of the car is $3 0^{\circ}$ ; 6 seconds later it is $6 0^{\circ}$ . Find the speed of the car and the further time it takes to reach the foot. Use $3 = 1.732$ .
Hint
Find both distances from the foot, subtract to get the distance covered in 6 s.
Answer
By alternate angles, the first distance is $\frac{30}{t a n 3 0 ^{\circ}} = 303$ m and the second is $\frac{30}{t a n 6 0 ^{\circ}} = 103$ m. Distance covered in 6 s $= 203$ m, so speed $= \frac{20 3}{6} = \frac{10 3}{3} \approx 5.77$ m/s. Further time $= 103 \div \frac{10 3}{3} = 3$ s.
Q2 · 3 marks · Short answerHarder board level
From a window 15 m above a level street, the angle of elevation of the top of the house opposite is $3 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Find the height of the house and the width of the street. Use $3 = 1.732$ .
Hint
The depression of the foot gives the street width; the horizontal through the window splits the house into two parts.
Answer
Street width $= \frac{15}{t a n 4 5 ^{\circ}} = 15$ m. Part of the house above the window level $= 15 tan 3 0^{\circ} = \frac{15}{3} = 53$ m. Height $= 15 + 53 \approx 15 + 8.66 = 23.66$ m.
Q3 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the top of a lighthouse is 60 m above sea level. A boat sailing straight towards it at a uniform speed is first seen at an angle of depression of $3 0^{\circ}$ . Exactly 2 minutes later, the angle of depression is $6 0^{\circ}$ .
(i) Find the two distances of the boat from the foot of the lighthouse. (1 mark)
(ii) Find the speed of the boat in m/min. (1 mark)
(iii) How much longer will the boat take to reach the foot, and how far from the foot will it be 30 seconds after the second sighting? Use $3 = 1.732$ . (2 marks)
Hint
The speed is not given; find it from the distance covered in 2 minutes.
Answer
(i) $\frac{60}{t a n 3 0 ^{\circ}} = 603$ m and $\frac{60}{t a n 6 0 ^{\circ}} = 203$ m. (1)
(ii) Distance covered $= 403$ m in 2 min, so speed $= 203 \approx 34.64$ m/min. (1)
(iii) Remaining $203$ m takes $\frac{20 3}{20 3} = 1$ minute. In 30 s it covers $103$ m, so it is $203 - 103 = 103 \approx 17.32$ m from the foot. (2)
Q4 · 5 marks · Long answerHarder board level
A man on the deck of a ship is 12 m above the water. He sees the top of a cliff at an angle of elevation of $6 0^{\circ}$ and its base at an angle of depression of $3 0^{\circ}$ . Find the distance of the cliff from the ship and the height of the cliff.
Hint
Depression with 12 m gives the distance; elevation with that distance gives the part above his eye.
Answer
Distance $= 123$ m (about 20.78 m); part above eye $= 123 \times 3 = 36$ m; cliff $= 36 + 12 = 48$ m.
Q5 · 3 marks · Short answerBoard level
Find the error. From a terrace 20 m high, the angle of depression of a car is $3 0^{\circ}$ . A student marks $3 0^{\circ}$ between the wall and the line of sight and writes distance $= 20 tan 3 0^{\circ} = \frac{20}{3}$ m. Explain the error and find the correct distance.
Hint
The angle with the wall is $9 0^{\circ}$ minus the depression.
Answer
Depression is measured from the horizontal, so the angle at the car is $3 0^{\circ}$ and the angle with the wall is $6 0^{\circ}$ . Correct distance $= \frac{20}{t a n 3 0 ^{\circ}} = 203$ m (about 34.64 m).
Q6 · 1 mark · MCQBoard level
From the top of a tower 30 m high on level ground, the angle of depression of a point on the ground is $6 0^{\circ}$ . The length of the line of sight from the top of the tower to the point is
(a) 60 m
(b) $303$ m
(c) $203$ m
(d) $103$ m
Hint
The line of sight is the hypotenuse: use sin of the angle at the point.
Answer
(c) $203$ m, since $\frac{30}{s i n 6 0 ^{\circ}} = \frac{60}{3} = 203$ .
Q7 · 1 mark · MCQBoard level
From the top of a tower on level ground, the angle of depression of a car changes from $3 0^{\circ}$ to $6 0^{\circ}$ as it drives straight towards the foot. The ratio of the car's first distance from the foot to its second distance is
(a) 1 : 3
(b) 3 : 1
(c) $3$ : 1
(d) 2 : 1
Hint
Write both distances in terms of the tower's height h.
Answer
(b) 3 : 1, since $3 h : \frac{h}{3} = 3 : 1$ .
Q8 · 3 marks · Short answerHarder board level
The line of sight from the top of a tower to a car on level ground is 60 m long, and the angle of depression of the car is $3 0^{\circ}$ . The car then drives straight towards the tower until the angle of depression is $6 0^{\circ}$ . Find the height of the tower and the distance the car moved. Use $3 = 1.732$ .
Hint
Use $sin 3 0^{\circ}$ and $cos 3 0^{\circ}$ with the line of sight first.
Answer
Height $= 60 sin 3 0^{\circ} = 30$ m; first distance $= 60 cos 3 0^{\circ} = 303$ m. Second distance $= \frac{30}{t a n 6 0 ^{\circ}} = 103$ m. Distance moved $= 203 \approx 34.64$ m.
Q9 · 5 marks · Long answerHarder board level
From the top A of a building AB, the angles of depression of the top C and the foot D of a tower CD, 20 m high and standing on the same level ground, are $4 5^{\circ}$ and $6 0^{\circ}$ . Find the height of the building and the distance between the building and the tower. Use $3 = 1.732$ .
Hint
Let the building be H m. Draw CE perpendicular to AB; then AE = H − 20 and CE = BD. Write BD from both triangles and equate.
Answer
$B D = \frac{H}{3}$ and $B D = H - 20$ , so $H (1 - \frac{1}{3}) = 20$ , giving $H = \frac{20 3}{3 - 1} = 30 + 103 \approx 47.32$ m; distance $= 10 (3 + 1) \approx 27.32$ m.
Q10 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a drone hovers at a fixed height h metres above a straight level road and films a parked van on the road. At first, the angle of depression of the van from the drone is $3 0^{\circ}$ . The drone then flies 40 m horizontally towards the van at 5 m/s, staying at the same height, and the angle of depression becomes $4 5^{\circ}$ .
(i) Write the first and the new horizontal distances between the drone and the van in terms of h. (1)
(ii) Find the height h of the drone. Use $3 = 1.732$ . (1)
(iii) How long did the 40 m take, and how much longer, at the same speed, will the drone take to be directly above the van? (2)
Hint
Transfer both angles to the van by alternate angles; the difference of the two horizontal distances is 40 m.
Answer
(i) $\frac{h}{t a n 3 0 ^{\circ}} = 3 h$ m and $\frac{h}{t a n 4 5 ^{\circ}} = h$ m. (1)
(ii) $3 h - h = 40$ , so $h = \frac{40}{3 - 1} = 20 (3 + 1) \approx 54.64$ m. (1)
(iii) $40 \div 5 = 8$ s. The drone is still $h \approx 54.64$ m from being above the van, so it needs $54.64 \div 5 \approx 10.93$ s more. (2)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define angle of depression.Show answer
Angle between the horizontal through the eye and the line of sight to an object below eye level.
2.Why does depression equal elevation?Show answer
Alternate interior angles: the horizontal at the eye is parallel to the ground.
3.Distance from foot for height h and depression $4 5^{\circ}$ ?Show answer
Equal to h.
4.Bigger depression means the object is...Show answer
Closer to the foot.
5.Angle between wall and line of sight for depression $3 0^{\circ}$ ?Show answer
$6 0^{\circ}$

Angle of Depression: frequently asked questions

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

Elevation is measured upward from the horizontal when the object is above you. Depression is measured downward from the horizontal when the object is below you. Both are measured from the horizontal at the observer's eye.

Is the angle of depression always equal to the angle of elevation?

The angle of depression of B from A equals the angle of elevation of A from B when the ground is level. This follows from alternate angles between parallel lines.

Do I need to draw a figure in the board exam?

Yes. A neat labelled figure is expected in written answers and helps you avoid the most common mistakes, such as marking the angle at the wall.

Which angles can come in depression questions?

Only 30, 45 and 60 degrees, with at most two right triangles in a problem, as per the 2026-27 syllabus.