MathsCBSE Class 10 · Some Applications of Trigonometry

Drawing Figures and Solving Word Problems in Heights and Distances

Q: Do I have to draw a figure in the board exam?

Yes, in written answers a neat labelled figure is expected and often carries marks. It also helps you avoid errors with angles and distances.

Q: How do I know whether to use sin, cos or tan?

Look at which side you know and which side you want. Height and ground distance give tan, a slanted length with height gives sin, and a slanted length with ground distance gives cos.

Q: Can angles other than 30, 45 and 60 degrees come?

No. The 2026-27 syllabus limits angles of elevation and depression to 30, 45 and 60 degrees, and problems use at most two right triangles.

Q: Should I leave answers in surd form or as decimals?

Leave the surd form unless the question gives a value of root 3 or asks for a decimal answer. If a value is given, convert at the last step.

Updated 14 October 2026 · Fact-checked

Drawing figures and solving word problems means turning a heights and distances question into a neat labelled right-triangle figure, then choosing tan, sin or cos and using standard values at 30, 45 and 60 degrees. The board expects a correct figure, the right ratio, clear reasons for transferred angles, and a final answer with units.

The angle of depression at the top equals the angle at the car (alternate angles), so tan 30° = h / d in the right triangle.

Illustration drawn for QuestionBuddy.

Understand Drawing Figures and Solving Word Problems

Every heights and distances question is a word problem about a right triangle hidden in a real scene: a tower, a cliff, a kite or a ladder. Your first job is not calculation. It is translation. You read the sentence and turn it into a figure where the ground is a horizontal line, every tower or pole is a vertical line, and the line of sight is a slanted line from the observer's eye.

Why does this work? A vertical object on level ground always meets the ground at $9 0^{\circ}$ . So the object, the ground and the line of sight form a right triangle, and the trigonometric ratios from Chapter 8 apply directly. Once the figure is right, the maths is usually one or two lines per triangle.

A good figure answers three questions before you write any ratio:

Where is the right angle? (At the foot of the vertical object.)
Where is the given angle? (At the observer's eye, measured from the horizontal.)
Which side is known and which is wanted? (This decides $tan$ , $sin$ or $cos$ .)

Three adjustments catch most students out. First, an angle of depression is drawn at the top, between the horizontal and the line of sight going down, and is then transferred to the ground point using alternate angles, because the horizontal at the eye is parallel to the ground. Second, if the observer has an eye height $p$ , the horizontal line of sight starts at height $p$ , so the triangle's vertical side is (object height $- p$ ). Third, when there are two triangles, the figure shows whether distances add (opposite sides of the object) or subtract (same side).

This topic ties together everything else in the chapter: elevation, depression, one-triangle and two-triangle problems all rely on the same figure routine. The syllabus keeps you within safe limits: at most two right triangles in a problem, and only the angles $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ . So the ratio you get will always be one of a few standard values, and surds like $3$ will appear often. Keep exact surd values until the last line, and use $3 = 1.732$ only if the question asks for a decimal answer.

Drawing Figures and Solving Word Problems: formulas to remember

Tangent in a heights figure: $tan θ = \frac{height}{horizontal distance}$; Use when the figure links the vertical side and the ground distance. This is the most used ratio.
Sine in a heights figure: $sin θ = \frac{height}{slanted length}$; Use for strings, ropes, ladders or a line of sight when the height is involved.
Cosine in a heights figure: $cos θ = \frac{horizontal distance}{slanted length}$; Use when the ground distance and a slanted length are involved.
Standard tan values: $tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3$; Only these three angles appear in the 2026-27 syllabus.
Standard sin and cos values: $sin 3 0^{\circ} = cos 6 0^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = cos 3 0^{\circ} = \frac{3}{2}, sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$; Needed when a slanted length is in the problem.
Transferring an angle of depression: $∠ depression at top = ∠ elevation at ground point$; Reason to write: alternate angles, since the horizontal at the eye is parallel to the ground.
Observer with eye height p: $H = p + d tan θ$; Solve the triangle above eye level, then add the eye height back.
Same side: distances subtract: $d = h cot α - h cot β, α < β$; Two observation points on the same side of the object; the larger angle is at the nearer point.
Opposite sides: distances add: $total = h cot α + h cot β$; Two points on opposite sides of the object, in line with its foot.
Rationalising: $\frac{1}{3 - 1} = \frac{3 + 1}{2}$; Appears often in 30° and 45° two-triangle problems.

How to solve Drawing Figures and Solving Word Problems questions

Use this same routine on every heights and distances word problem. It keeps the figure correct, and a correct figure makes the calculation short.

1Read the question once fully and list the objects (tower, observer, boats), every given length and every given angle.
2Draw the ground as a horizontal line and each object as a vertical line. Label every point with a capital letter and mark the right angle at each foot.
3Mark the observer's eye. If an eye height is given, draw the horizontal from eye level, not the ground. For a depression angle, draw the horizontal at the top, mark the angle there, then transfer it to the ground point and write 'alternate angles'.
4Write known lengths on the figure and name unknowns $h$ and $x$ . Decide from the figure whether two ground distances add (opposite sides) or subtract (same side).
5In each triangle, pick the ratio containing one known side and the wanted side: height and ground give $tan$ ; a slanted length gives $sin$ or $cos$ .
6Substitute the standard value, solve, and with two triangles solve the pair of equations. Rationalise any surd in the denominator.
7Add back any eye height or building height the question needs, convert to decimals only if asked, and write units.
8Check: the larger angle should match the shorter ground distance, and at $4 5^{\circ}$ height equals distance.

Drawing Figures and Solving Word Problems 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.

Drawing the figure and modelling a word problem

Suppose at a kite festival on a flat school ground you hold the kite string 1 m above the ground. You have let out 80 m of string, which is taut and straight, and it makes an angle of $3 0^{\circ}$ with the horizontal. A gust of wind then lifts the kite until the same 80 m of string makes $6 0^{\circ}$ with the horizontal.

How far does the kite shift horizontally, and is it towards you or away from you? What is the kite's new height above the ground? You then let out more string, keeping the angle at $6 0^{\circ}$ , until the kite is 100 m above the ground. How much extra string did you let out? Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Your hand is 1 m above the ground, so draw the horizontal from the hand. The string is the hypotenuse in both positions.
First position: horizontal distance $= 80 cos 3 0^{\circ} = 80 \times \frac{3}{2} = 403$ m.
Second position: horizontal distance $= 80 cos 6 0^{\circ} = 80 \times \frac{1}{2} = 40$ m.
The horizontal distance has decreased, so the kite has moved towards you.
Horizontal shift $= 403 - 40 = 40 (3 - 1) = 40 \times 0.732 = 29.28$ m.
New height above hand $= 80 sin 6 0^{\circ} = 80 \times \frac{3}{2} = 403 = 69.28$ m.
New height above ground $= 69.28 + 1 = 70.28$ m.
For a height of 100 m above the ground, the kite must be $100 - 1 = 99$ m above your hand.
String needed $= \frac{99}{s i n 6 0 ^{\circ}} = \frac{198}{3} = 663 = 66 \times 1.732 = 114.31$ m.
Extra string $= 114.31 - 80 = 34.31$ m.

Answer: The kite moves $40 (3 - 1) \approx 29.28$ m horizontally towards you, and its new height is about 70.28 m above the ground. To reach 100 m at $6 0^{\circ}$ you need $663 \approx 114.31$ m of string, so you let out about 34.31 m more.

Memory hook: Same string, bigger angle: the kite climbs and comes closer; cos gives the ground, sin gives the height.

Tricks to remember Drawing Figures and Solving Word Problems

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

H-V-S before any ratio
Draw the Horizontal ground, the Vertical object, then the Slanted line of sight. Only then mark the angle at the eye.
Use it when: Stops you from writing a ratio before the triangle is clear.
Depression drops down
The depression angle sits up top outside the triangle; it drops down to the ground point by alternate angles.
Use it when: Prevents putting the depression angle at the top vertex inside the triangle.
Same side subtract, opposite add
Points on the same side of a tower: the far distance minus the near distance. Points on opposite sides: add them.
Use it when: Sets up two-triangle problems correctly.
Big angle, close point
A larger angle of elevation means you are nearer the object.
Use it when: A quick check that your figure's labels are not swapped.

Drawing Figures and Solving Word Problems: solved examples

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

Example 1: A ladder that slips · 3 marks · Short answerHarder board level

A ladder 6 m long leans against a vertical wall on level ground, making an angle of $6 0^{\circ}$ with the ground. Its foot slips outwards until the ladder makes $4 5^{\circ}$ with the ground, with the top still on the wall. How far does the top slide down, and how far does the foot move? Use $2 = 1.414$ and $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: wall AB with foot B; first position of the ladder AC, second position DE with D on the wall and E on the ground. Both are right triangles with the right angle at B, and the ladder is the hypotenuse each time. (½ mark)
At $6 0^{\circ}$ : height $A B = 6 sin 6 0^{\circ} = 33 \approx 5.196$ m and foot $B C = 6 cos 6 0^{\circ} = 3$ m. (1 mark)
At $4 5^{\circ}$ : height $D B = 6 sin 4 5^{\circ} = 32 \approx 4.242$ m and foot $B E = 32 \approx 4.242$ m. (1 mark)
Top slides down $33 - 32 \approx 5.196 - 4.242 = 0.954$ m; foot moves out $32 - 3 \approx 1.242$ m. (½ mark)

Answer: The top slides down about 0.95 m and the foot moves out about 1.24 m.

Example 2: Elevation and depression from a window · 3 marks · Short answerBoard level

From a window 12 m above level ground, the angle of elevation of the top of a tower across the road is $6 0^{\circ}$ and the angle of depression of its foot is $4 5^{\circ}$ . Find the height of the tower. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: window W, with WG = 12 m vertical to the ground at G; tower PQ with Q its foot on the ground and P its top. Horizontal from W meets PQ at R, so RQ = WG = 12 m and WR = GQ. (½ mark)
Depression of Q from W is $4 5^{\circ}$ , so $∠ W QG = 4 5^{\circ}$ (alternate angles). In $△ W GQ$ : $tan 4 5^{\circ} = \frac{12}{GQ}$ , so $GQ = 12$ m and WR = 12 m. (1 mark)
In $△ P R W$ : $tan 6 0^{\circ} = \frac{P R}{W R}$ , so $P R = 123$ m. (1 mark)
$P Q = P R + R Q = 123 + 12 = 12 (1.732 + 1) = 32.784$ m. (½ mark)

Answer: Height of the tower $= 12 (3 + 1) \approx 32.78$ m.

Example 3: A tower on a building · 3 marks · Short answerBoard level

A mobile tower 10 m tall stands on the flat roof of a building. From a point P on level ground, the angles of elevation of the bottom and the top of the tower are $4 5^{\circ}$ and $6 0^{\circ}$ . Find the height of the building and the distance of P from it. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: building BC with foot C on the ground and top B; tower AB on top, AB = 10 m. P on the ground, PC = d. Both triangles PCB and PCA are right-angled at C. Let BC = b. (½ mark)
In $△ P C B$ : $tan 4 5^{\circ} = \frac{b}{d}$ , so $d = b$ . (½ mark)
In $△ P C A$ : $tan 6 0^{\circ} = \frac{b + 10}{d}$ , so $b + 10 = 3 b$ . (1 mark)
$b (3 - 1) = 10$ , so $b = \frac{10}{3 - 1} = 5 (3 + 1) \approx 13.66$ m, and $d = b \approx 13.66$ m. The two heights are subtracted: tower = (building + tower) − building. (1 mark)

Answer: The building is $5 (3 + 1) \approx 13.66$ m high, and P is about 13.66 m from it.

Example 4: Moving observer with eye height · 5 marks · Long answerBoard level

A girl whose eyes are 1.5 m above the ground stands on level ground in front of a vertical building. From her eyes, the angle of elevation of the top of the building is $3 0^{\circ}$ . She walks 20 m straight towards the building, and the angle of elevation becomes $4 5^{\circ}$ . Find the height of the building and how far she now is from the building. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: building PQ, top P, foot Q. Her eye starts at A and moves to B, both 1.5 m above ground; the horizontal through A and B meets PQ at R, so RQ = 1.5 m. AB = 20 m, $∠ P A R = 3 0^{\circ}$ , $∠ P B R = 4 5^{\circ}$ . Let PR = h, BR = x, so AR = x + 20 (same side: distances differ by 20). (1 mark)
In $△ P B R$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $h = x$ . (1 mark)
In $△ P A R$ : $tan 3 0^{\circ} = \frac{h}{x + 20}$ , so $3 h = x + 20 = h + 20$ . (1 mark)
$h (3 - 1) = 20$ , so $h = \frac{20}{3 - 1} = \frac{20 ( 3 + 1 )}{2} = 10 (3 + 1) = 27.32$ m. (1 mark)
Building height $= P R + R Q = 27.32 + 1.5 = 28.82$ m; her distance from the building $= x = h = 27.32$ m. (1 mark)

Answer: Height of the building $\approx 28.82$ m; she is now about 27.32 m from it.

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

A forest department builds a vertical fire-watch tower 30 m high on level ground. From the top of the tower, a guard sees a campsite B on the ground, in line with the foot of the tower, at an angle of depression of $3 0^{\circ}$ . A ranger leaves B and walks in a straight line towards the foot of the tower at a steady 1.5 m/s. After some time, the guard sees the ranger at an angle of depression of $4 5^{\circ}$ .
(i) Find the distance of campsite B from the foot of the tower. (1 mark)
(ii) How far is the ranger from the foot of the tower when the angle of depression is $4 5^{\circ}$ ? (1 mark)
(iii) How long does the angle of depression take to change from $3 0^{\circ}$ to $4 5^{\circ}$ , and how many more seconds does the ranger need to reach the foot of the tower? (2 marks)
Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: tower TF, T the top, F the foot, TF = 30 m; B on the ground, and C the ranger's position when the angle is $4 5^{\circ}$ , with C between B and F (larger angle, nearer point). By alternate angles, $∠ T B F = 3 0^{\circ}$ and $∠ T C F = 4 5^{\circ}$ .
(i) In $△ T F B$ : $tan 3 0^{\circ} = \frac{30}{B F}$ , so $B F = 303 = 51.96$ m. (1 mark)
(ii) In $△ T F C$ : $tan 4 5^{\circ} = \frac{30}{C F}$ , so $C F = 30$ m. (1 mark)
(iii) Same side, so subtract: $B C = 303 - 30 = 30 (3 - 1) = 30 \times 0.732 = 21.96$ m. (½ mark)
Time from B to C $= \frac{21.96}{1.5} = 14.64$ s. (½ mark)
Remaining distance CF = 30 m, so extra time $= \frac{30}{1.5} = 20$ s. Check: total time $= 34.64$ s and $34.64 \times 1.5 = 51.96$ m, which equals BF. (1 mark)

Answer: (i) $303 \approx 51.96$ m (ii) 30 m (iii) about 14.64 s for the angle to change from $3 0^{\circ}$ to $4 5^{\circ}$ , then 20 s more to reach the foot.

Where students lose marks in Drawing Figures and Solving Word Problems

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

Mistake: Placing the angle of depression inside the triangle at the top
The depression is measured from the horizontal at the eye, which lies outside the triangle.
Wrong working
Cliff 20 m, angle of depression $3 0^{\circ}$ : writes $tan 3 0^{\circ} = \frac{x}{20}$ , so $x = \frac{20}{3}$ m.
Right working
Angle at the boat $= 3 0^{\circ}$ (alternate angles). $tan 3 0^{\circ} = \frac{20}{x}$ , $x = 203$ m.
Do this: Draw the horizontal at the top, mark the depression there, transfer it to the ground and write 'alternate angles'.
Usually costs: Usually most of the marks for that triangle.
Mistake: Ignoring the observer's eye height
The horizontal line of sight starts at eye level, so the triangle's vertical side is shorter than the object.
Wrong working
Eyes 1.5 m high, 20 m from a building, elevation $4 5^{\circ}$ : height $= 20 tan 4 5^{\circ} = 20$ m.
Right working
Height above eye $= 20$ m, so building $= 20 + 1.5 = 21.5$ m.
Do this: Draw a dotted horizontal from the eye and add the eye height at the end.
Usually costs: The final answer mark, sometimes more.
Mistake: Adding distances when points are on the same side
On the same side, the far distance includes the near one, so the gap is a difference.
Wrong working
Boats at 75 m and $753$ m on the same side: gap $= 75 + 753$ m.
Right working
Gap $= 753 - 75 = 75 (3 - 1) \approx 54.9$ m.
Do this: Look at the figure: if both points are on one side of the foot, subtract.
Usually costs: 1 to 2 marks.
Mistake: Using tan when a slanted length is given
Strings, ladders and lines of sight are hypotenuses; tan does not contain the hypotenuse.
Wrong working
String 100 m at $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{h}{100}$ , $h = \frac{100}{3}$ m.
Right working
$sin 3 0^{\circ} = \frac{h}{100}$ , $h = 50$ m.
Do this: Label each side opposite, adjacent or hypotenuse before choosing the ratio.
Usually costs: Most marks for the question.

Exam tips

Always draw and label the figure in a written answer; examiners award marks for a correct figure, and it shows them your method.
Write the reason 'alternate angles' every time you transfer an angle of depression.
Keep surds exact until the last line, and use $3 = 1.732$ only when the question asks for a decimal value.
In case studies, draw one figure for the whole case and reuse earlier part answers in later parts.
Before finishing, check that the larger angle sits at the nearer point and write the unit in the final answer.

Practice questions on Drawing Figures and Solving Word Problems

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

Q1 · 3 marks · Short answerHarder board level
A girl whose eyes are 1.5 m above level ground stands 18 m from a building. From her eyes, the angle of elevation of the foot of a flagstaff fixed on the roof is $3 0^{\circ}$ , and that of the top of the flagstaff is $4 5^{\circ}$ . Find the height of the building and the length of the flagstaff. Use $3 = 1.732$ .
Hint
Draw the horizontal through her eyes. Both right triangles share the 18 m side; add 1.5 m only to the building.
Answer
Part of the building above eye level $= 18 tan 3 0^{\circ} = 63 \approx 10.39$ m, so the building is $1.5 + 63 \approx 11.89$ m high. Top of flagstaff above eye level $= 18 tan 4 5^{\circ} = 18$ m. Flagstaff $= 18 - 63 \approx 7.61$ m.
Q2 · 3 marks · Short answerHarder board level
A car moves at a uniform 18 km/h along a straight level road towards the foot of a vertical tower. From the top, the angle of depression of the car changes from $3 0^{\circ}$ to $6 0^{\circ}$ in 12 seconds. Find the height of the tower and how many more seconds the car takes to reach the foot. Use $3 = 1.732$ .
Hint
Convert the speed to m/s first; the height is the unknown.
Answer
Speed $= 18 \times \frac{5}{18} = 5$ m/s, so the car covers 60 m in 12 s. With height $h$ , the distances are $3 h$ and $\frac{h}{3}$ , so $3 h - \frac{h}{3} = \frac{2 h}{3} = 60$ and $h = 303 \approx 51.96$ m. Remaining distance $= \frac{h}{3} = 30$ m, which takes 6 more seconds.
Q3 · 5 marks · Long answerHarder board level
A mobile tower stands on the flat roof of a building. From a point P on level ground, the angle of elevation of the top of the building is $3 0^{\circ}$ and the angle of elevation of the top of the tower is $6 0^{\circ}$ . The tower is 12 m taller than the building. Find (i) the height of the building and of the tower, (ii) the distance of P from the foot of the building, (iii) the length of the line of sight from P to the top of the tower. Use $3 = 1.732$ .
Hint
Let the building be b, the tower t and the ground distance d. Write d from the $3 0^{\circ}$ triangle, use it in the $6 0^{\circ}$ triangle and find a relation between t and b first.
Answer
$tan 3 0^{\circ} = \frac{b}{d}$ gives $d = 3 b$ . $tan 6 0^{\circ} = \frac{b + t}{d}$ gives $b + t = 3 d = 3 b$ , so $t = 2 b$ . With $t = b + 12$ : $2 b = b + 12$ , so $b = 12$ m and $t = 24$ m. (ii) $d = 123 \approx 20.78$ m. (iii) Line of sight $= \frac{36}{s i n 6 0 ^{\circ}} = \frac{72}{3} = 243 \approx 41.57$ m. Check: $cos 6 0^{\circ} = \frac{12 3}{24 3} = \frac{1}{2}$ .
Q4 · 5 marks · Long answerBoard level
From the top of a vertical tower, the angles of depression of two points on level ground, on opposite sides of the tower and in one straight line with its foot, are $3 0^{\circ}$ and $6 0^{\circ}$ . The points are 100 m apart. Find the height of the tower, the distance of each point from its foot, and the length of the line of sight from the top to the nearer point. Use $3 = 1.732$ .
Hint
Write both ground distances in terms of h, add them and set the sum to 100.
Answer
Distances: $h 3$ (for $3 0^{\circ}$ ) and $\frac{h}{3}$ (for $6 0^{\circ}$ ). $h 3 + \frac{h}{3} = 100$ gives $\frac{4 h}{3} = 100$ , so $h = 253 \approx 43.3$ m. Distances: 75 m and 25 m. Line of sight to nearer point $= \frac{h}{s i n 6 0 ^{\circ}} = \frac{25 3}{3 /2} = 50$ m.
Q5 · 3 marks · Short answerHOTS / competency
Question: From the top of a vertical tower 60 m high, the angles of depression of two cars on a straight level road, on the same side of the tower and in line with its foot, are $3 0^{\circ}$ and $6 0^{\circ}$ . Find the distance between the cars.
A student wrote: Car at $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{60}{x}$ , so $x = 203$ m. Car at $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{60}{y}$ , so $y = 603$ m. Distance between cars $= x + y = 803 \approx 138.56$ m.
(i) Identify the error and explain why it is an error. (ii) Write the full correct solution. (iii) For what arrangement of the cars would the student's answer be correct? Use $3 = 1.732$ .
Hint
Draw both cars on one side of the foot. Does the far distance already contain the near one?
Answer
(i) Both distances from the foot are correct, but the student added them. On the same side, the far car's distance FB already includes the near car's distance FA, so the gap is FB − FA. (ii) Tower TF = 60 m; near car A, far car B; by alternate angles $∠ T A F = 6 0^{\circ}$ and $∠ T B F = 3 0^{\circ}$ . $F A = \frac{60}{3} = 203$ m, $F B = 603$ m, so $A B = 603 - 203 = 403 \approx 69.28$ m. (iii) The sum $803$ m is correct only if the cars are on opposite sides of the tower.
Q6 · 1 mark · MCQBoard level
From the top of a vertical tower 20 m high, the angles of depression of two cars on the same side of the tower, in line with its foot, are $4 5^{\circ}$ and $6 0^{\circ}$ . The distance between the cars is:
(a) $\frac{20 ( 3 - 3 )}{3}$ m
(b) $20 (3 - 1)$ m
(c) $\frac{20 ( 3 + 3 )}{3}$ m
(d) $203$ m
Hint
Same side: subtract the two distances from the foot.
Answer
(a). Distances are $\frac{20}{t a n 4 5 ^{\circ}} = 20$ m and $\frac{20}{t a n 6 0 ^{\circ}} = \frac{20}{3} = \frac{20 3}{3}$ m, so the gap is $20 - \frac{20 3}{3} = \frac{20 ( 3 - 3 )}{3} \approx 8.45$ m. Option (c) is the trap of adding.
Q7 · 1 mark · MCQBoard level
A boy whose eyes are 1.5 m above level ground stands 30 m from a vertical tower and sees its top at an angle of elevation of $6 0^{\circ}$ . The height of the tower is:
(a) $(303 + 1.5)$ m
(b) $303$ m
(c) $(103 + 1.5)$ m
(d) $31.5$ m
Hint
The 1.5 m is added after using tan.
Answer
(a). Height above his eyes $= 30 tan 6 0^{\circ} = 303$ m; add the eye height: $(303 + 1.5)$ m $\approx 53.46$ m.
Q8 · 3 marks · Short answerHarder board level
Two vertical poles of equal height stand on opposite banks of a straight canal 120 m wide, exactly facing each other. From a point on the line joining their feet, the angles of elevation of the tops of the poles are $3 0^{\circ}$ and $4 5^{\circ}$ . Find the height of the poles and the distance of the point from each pole. A boat then crosses from the point to the nearer pole at 2 m/s; how long does it take? Use $3 = 1.732$ .
Hint
The larger angle belongs to the nearer pole. The two distances add to 120 m.
Answer
Let the height be $h$ . Distances: $\frac{h}{t a n 3 0 ^{\circ}} = 3 h$ and $\frac{h}{t a n 4 5 ^{\circ}} = h$ . So $3 h + h = 120$ and $h = \frac{120}{3 + 1} = 60 (3 - 1) \approx 43.92$ m. The point is about 43.92 m from the nearer pole and $120 - 43.92 = 76.08$ m from the other. Time $= 43.92 \div 2 \approx 21.96$ s.
Q9 · 1 mark · Assertion-reasonHarder board level
Assertion (A): From the top of a vertical tower of height h, the angles of depression of two points on level ground, on opposite sides of the tower and in line with its foot, are $3 0^{\circ}$ and $6 0^{\circ}$ . The distance between the points is $\frac{4 h}{3}$ .
Reason (R): For two points on the same side of the tower, with angles of depression $3 0^{\circ}$ and $6 0^{\circ}$ , the distance between them is $h (cot 3 0^{\circ} - cot 6 0^{\circ})$ .
(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.
Hint
Work out A on its own. Does it use the same-side rule?
Answer
(b). A: the distances from the foot are $h cot 3 0^{\circ} = 3 h$ and $h cot 6 0^{\circ} = \frac{h}{3}$ ; on opposite sides they add to $\frac{3 h + h}{3} = \frac{4 h}{3}$ , so A is true. R is also true (same side: subtract), but A uses the opposite-side rule (add), so R does not explain A.

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.From which line is an angle of elevation or depression measured?Show answer
From the horizontal line through the observer's eye.
2.Which ratio links height and ground distance?Show answer
$tan θ = \frac{height}{ground distance}$ .
3.What is $tan 6 0^{\circ}$ ?Show answer
$3$ .
4.Why does the angle of depression equal the angle at the ground point?Show answer
Alternate angles: the horizontal at the eye is parallel to the ground.
5.At $4 5^{\circ}$ , how do height and ground distance compare?Show answer
They are equal.

Drawing Figures and Solving Word Problems: frequently asked questions

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

Yes, in written answers a neat labelled figure is expected and often carries marks. It also helps you avoid errors with angles and distances.

How do I know whether to use sin, cos or tan?

Look at which side you know and which side you want. Height and ground distance give tan, a slanted length with height gives sin, and a slanted length with ground distance gives cos.

Can angles other than 30, 45 and 60 degrees come?

No. The 2026-27 syllabus limits angles of elevation and depression to 30, 45 and 60 degrees, and problems use at most two right triangles.

Should I leave answers in surd form or as decimals?

Leave the surd form unless the question gives a value of root 3 or asks for a decimal answer. If a value is given, convert at the last step.