MathsCBSE Class 10 · Some Applications of Trigonometry

Problems with Two Right Triangles in Heights and Distances

Q: How do I know which side the two triangles share?

Look for the length that belongs to both triangles in your figure. When one tower is seen from two ground points, the tower's height is shared. When one observer looks up and down, the horizontal distance is shared.

Q: Can a board question have three right triangles?

No. The 2026-27 syllabus says problems should not involve more than two right triangles, and the angles are limited to 30, 45 and 60 degrees.

Q: Should I give the answer in surd form or as a decimal?

Give the exact surd form unless the question gives a value such as root 3 = 1.732 or asks for a decimal. If a value is given, show the surd form first, then substitute.

Q: Do I need to draw the figure if marks are not mentioned for it?

Yes. Examiners usually give a mark for a correct figure in long answers, and the figure tells you whether to add or subtract distances.

Q: Is a formula like h = d divided by (cot A minus cot B) accepted?

It is safer to write both tan equations and solve them, because step marks are given for each equation. Use the formula only to check your final answer.

Updated 14 October 2026 · Fact-checked

A two right triangle problem gives two angles of elevation or depression, so the figure has two right triangles sharing a side, usually a tower's height or a ground distance. You write one tan equation for each triangle and solve them together. The board sets these as 3-mark short answers, 5-mark long answers and case studies.

Both triangles share the tower height h, so the two tan equations can be solved together, with distances x and x + d.

Illustration drawn for QuestionBuddy.

Understand Problems with Two Right Triangles

In a one-triangle problem you know one side and one angle, so a single trigonometric ratio finishes the job. In a two right triangle problem you are given two angles, and there is usually not enough information to solve either triangle alone. Each triangle has one known angle but its sides are unknown. The two triangles together give enough information.

The trick is the shared side. Two triangles in a heights and distances figure almost always share something. It may be the tower, seen from two points on the ground. It may be the horizontal distance between a building and a tower, seen by one observer looking up and down. Because the side is shared, the same letter, such as $h$ or $x$ , appears in both equations. Two equations in two unknowns can then be solved, exactly as in pair of linear equations.

There are four common layouts. Same side: two points on one straight line through the foot of a tower; the far distance is the near distance plus the gap, so you work with $x$ and $x + d$ . Opposite sides: the tower or the observer sits between the two points, so the two horizontal distances add to the total. Up and down from a building: one horizontal line through the observer's eye splits the taller object into two parts, and the ground distance is common. Object on an object: a flagstaff on a building, seen from one ground point; the ground distance is common and the heights differ by the flagstaff's length.

This builds on everything earlier in the chapter. You still measure every angle from the horizontal, you still transfer an angle of depression to the ground using alternate angles, and you still pick $tan$ whenever only heights and ground distances appear. The new skill is deciding which side is shared and writing both distances in terms of the same unknowns.

The conditions in the 2026-27 syllabus are strict: at most two right triangles, and the angles are only $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ . So your answers usually come out in terms of $3$ , or with $3 - 1$ or $3 + 1$ in a denominator that you rationalise. A useful self-check: on the same side, the larger angle belongs to the nearer point.

Problems with Two Right Triangles: formulas to remember

Tangent in each triangle: $tan θ = \frac{height}{horizontal distance}$; Write one such equation for each of the two triangles.
Same side, two points d apart: $h = \frac{d}{c o t α - c o t β}, α < β$; α is the angle at the farther point. Use only to check; show both equations in the exam.
Same side, 30° and 60°: $h = \frac{3}{2} d, nearer distance = \frac{d}{2}$; Quick check for the most common pair of angles.
Same side, 30° and 45°: $h = \frac{d}{3 - 1} = \frac{d ( 3 + 1 )}{2}$; Rationalise by multiplying by the conjugate √3 + 1.
Opposite sides: $total distance = h (cot α + cot β)$; Distances on opposite sides of the vertical line add.
Elevation and depression from a building of height b: $H = b + b cot β tan α$; β is the depression of the foot, α the elevation of the top. Distance between them is b cot β.
Flagstaff of length f on a building: $f = d (tan β - tan α)$; d is the common ground distance, α and β the elevations of the bottom and top of the flagstaff.

How to solve Problems with Two Right Triangles questions

Use this routine on every two-triangle question. The figure and the shared side do most of the work.

1Draw the ground horizontally and the tower, building or pole vertically. Label every point with a capital letter.
2Mark both angles at the correct points, measured from the horizontal. Transfer any angle of depression to the ground using alternate angles and write the reason.
3Identify the shared side (height or ground distance) and call it $h$ or $x$ . Write the other lengths in terms of it: $x + d$ for the same side, $d - x$ for opposite sides.
4Write one $tan$ equation for each right triangle, naming the triangle each time.
5Solve the two equations together: express $h$ from both and equate, or substitute.
6Rationalise any surd denominator, then use $3 = 1.732$ only if the question asks for a decimal.
7Check: the larger angle should belong to the nearer point, and substituting back should satisfy both equations. Write the final answer with units.

Problems with Two Right Triangles 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.

Problems with Two Right Triangles

Imagine your colony puts up a tall vertical light pole for a festival on a level open ground. You and your friend stand on opposite sides of the pole, in one straight line with its foot, 30 m apart. You see the top of the pole at an angle of elevation of $4 5^{\circ}$ and your friend sees it at $6 0^{\circ}$ . Ignore your heights.

Find the height of the pole and how far each of you is from its foot. Give the exact height and its value using $3 = 1.732$ .

Try it first, then show the solutionSolution

Let the pole be PQ with top P, foot Q, PQ = h. You stand at A, your friend at B, with Q between you and AB = 30 m.
In right $△ P Q A$ : $tan 4 5^{\circ} = \frac{h}{A Q}$ , so $A Q = h$ .
In right $△ P QB$ : $tan 6 0^{\circ} = \frac{h}{B Q}$ , so $B Q = \frac{h}{3}$ .
Opposite sides, so the distances add: $h + \frac{h}{3} = 30$ , so $h (3 + 1) = 303$ .
$h = \frac{30 3 ( 3 - 1 )}{( 3 + 1 ) ( 3 - 1 )} = \frac{30 3 ( 3 - 1 )}{2} = 15 (3 - 3)$ m.
$h = 15 (3 - 1.732) = 15 \times 1.268 = 19.02$ m, so AQ = 19.02 m.
$B Q = 30 - 19.02 = 10.98$ m. Check: $\frac{19.02}{10.98} \approx 1.732 = tan 6 0^{\circ}$ .

Answer: Pole height $= 15 (3 - 3)$ m, about 19.02 m; you are about 19.02 m and your friend about 10.98 m from the foot.

Memory hook: Pole in the middle, friends on both sides: the two ground distances add to the gap between you.

Tricks to remember Problems with Two Right Triangles

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

Same side subtract, opposite sides add
If both points are on one side of the tower, the gap between them is the difference of their distances. If the tower is between them, the total is the sum.
Use it when: Decide this before writing any equation; it fixes $x + d$ or $d - x$ .
Big angle, near point
The steeper you look up, the closer you are. The larger angle always belongs to the nearer point.
Use it when: Check your figure before solving a same-side question.
Find the shared side first
Ask: what do both triangles have in common? That side gets one letter and appears in both equations.
Use it when: Every two-triangle question, especially building-and-tower and flagstaff types.
30-60 halves
For same-side points d apart with angles $3 0^{\circ}$ and $6 0^{\circ}$ , the near point is $\frac{d}{2}$ from the foot and the height is $\frac{3}{2} d$ .
Use it when: A fast check on MCQs and long answers.

Problems with Two Right Triangles: solved examples

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

Example 1: Two boats seen from a lighthouse · 3 marks · Short answerBoard level

From the top of a vertical lighthouse 75 m high, the angles of depression of two boats are $3 0^{\circ}$ and $6 0^{\circ}$ . Both boats are on the same side of the lighthouse, in one straight line with its foot. Find the distance between the boats. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let PQ be the lighthouse, P the top, Q the foot, PQ = 75 m. A is the farther boat, B the nearer boat, both on line QA. (½ mark for figure)
Angles of depression equal angles of elevation by alternate angles, so $∠ P A Q = 3 0^{\circ}$ and $∠ P B Q = 6 0^{\circ}$ . (½ mark)
In right $△ P QB$ : $tan 6 0^{\circ} = \frac{75}{B Q}$ , so $B Q = \frac{75}{3} = 253$ m. (1 mark)
In right $△ P Q A$ : $tan 3 0^{\circ} = \frac{75}{A Q}$ , so $A Q = 753$ m. (½ mark)
$A B = A Q - B Q = 503 = 50 \times 1.732 = 86.6$ m. (½ mark)

Answer: The boats are $503$ m, about 86.6 m, apart.

Example 2: A pole and a tower of different heights across a road · 5 marks · Long answerHarder board level

A vertical street-light pole AB and a vertical mobile tower CD stand on opposite sides of a straight road 30 m wide. Their feet B and D are directly opposite each other, and the tower is twice as tall as the pole. From a point P on the road, on the line BD between the feet, the angle of elevation of the top A of the pole is $3 0^{\circ}$ and that of the top C of the tower is $6 0^{\circ}$ . Find the distances of P from the feet of the pole and the tower, and the heights of both. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: B, P, D on one line with BD = 30 m. Let AB = h, so CD = 2h. Let BP = x, so PD = 30 - x. $∠ A P B = 3 0^{\circ}$ and $∠ C P D = 6 0^{\circ}$ . (1 mark for figure and unknowns)
In right $△ A B P$ : $tan 3 0^{\circ} = \frac{h}{x}$ , so $x = 3 h$ . (1 mark)
In right $△ C D P$ : $tan 6 0^{\circ} = \frac{2 h}{30 - x}$ , so $30 - x = \frac{2 h}{3}$ . (1 mark)
Add the two distances (opposite sides): $30 = 3 h + \frac{2 h}{3} = \frac{3 h + 2 h}{3} = \frac{5 h}{3}$ , so $h = \frac{30 3}{5} = 63$ m. (1 mark)
$B P = 3 \times 63 = 18$ m and $P D = 30 - 18 = 12$ m. Pole $= 63 \approx 10.39$ m, tower $= 123 \approx 20.78$ m. (½ mark)
Check: $\frac{6 3}{18} = \frac{1}{3} = tan 3 0^{\circ}$ and $\frac{12 3}{12} = 3 = tan 6 0^{\circ}$ . (½ mark)

Answer: P is 18 m from the pole and 12 m from the tower. Pole height $= 63$ m, about 10.39 m; tower height $= 123$ m, about 20.78 m.

Example 3: Tower from two points, 30° and 45° · 3 marks · Short answerBoard level

A and B are two points on level ground on the same side of a vertical tower PQ (P is the top, Q the foot), in one straight line with Q. AB = 50 m and B is nearer to the tower. The angles of elevation of P from A and B are $3 0^{\circ}$ and $4 5^{\circ}$ . Find the height of the tower and the distances of A and B from its foot. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Draw PQ vertical, Q, B, A in one line with B nearer. Let PQ = h and BQ = x, so AQ = x + 50. (½ mark for figure)
In right $△ P QB$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $h = x$ . (½ mark)
In right $△ P Q A$ : $tan 3 0^{\circ} = \frac{h}{x + 50}$ , so $3 h = x + 50$ . (½ mark)
Substitute $x = h$ : $3 h - h = 50$ , so $h = \frac{50}{3 - 1}$ . (½ mark)
Rationalise: $h = \frac{50 ( 3 + 1 )}{2} = 25 (3 + 1) = 25 \times 2.732 = 68.3$ m. (½ mark)
BQ = 68.3 m and AQ = 68.3 + 50 = 118.3 m. Check: $\frac{68.3}{118.3} \approx 0.577 = tan 3 0^{\circ}$ . (½ mark)

Answer: Height $= 25 (3 + 1)$ m, about 68.3 m; B is 68.3 m and A is 118.3 m from the foot.

Example 4: Looking up and down from a building · 3 marks · Short answerBoard level

From the top of a building 12 m high, the angle of elevation of the top of a vertical tower is $6 0^{\circ}$ and the angle of depression of the foot of the tower is $4 5^{\circ}$ . The building and tower stand on the same level ground. Find the height of the tower and the distance between them. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Let AB be the building (A top, B foot), AB = 12 m, and CD the tower (C top, D foot). Draw AE horizontal meeting CD at E, so ED = AB = 12 m and AE = BD. (½ mark for figure)
Angle of depression of D is $4 5^{\circ}$ , so $∠ A D B = 4 5^{\circ}$ (alternate angles). In right $△ A B D$ : $tan 4 5^{\circ} = \frac{12}{B D}$ , so BD = 12 m. (1 mark)
In right $△ A E C$ : $tan 6 0^{\circ} = \frac{C E}{A E} = \frac{C E}{12}$ , so $C E = 123$ m. (1 mark)
$C D = C E + E D = 123 + 12 = 12 (3 + 1) = 12 \times 2.732 = 32.78$ m. (½ mark)

Answer: Tower height $= 12 (3 + 1)$ m, about 32.78 m; distance between them = 12 m.

Example 5: Case study: width of a river · 4 marks · Case-basedHOTS / competency

A group of students on a nature camp wants to know the width of a river without crossing it. A vertical tower stands right at the edge of the far bank. Standing at a point on the near bank directly opposite the tower, a student finds the angle of elevation of the top of the tower is $4 5^{\circ}$ . She walks 20 m straight back from the river, along the line joining the foot of the tower to her first position, and the angle of elevation is now $3 0^{\circ}$ . Take the banks as level and ignore her height.
(i) Choose letters for the points and the unknown lengths, state which length is shared by the two right triangles, and write the equation you get from the first position. (1 mark)
(ii) Write the equation you get from the second position. (1 mark)
(iii) Find the width of the river and the height of the tower in rationalised surd form, and then as decimals. Use $3 = 1.732$ . (2 marks)

Try it first, then show the solutionSolution

(i) Let the tower be PQ (P top, Q foot), the first position A and the second B, with Q, A, B in one line. Let PQ = h and the width AQ = x. The height h is shared by both triangles. In right $△ P Q A$ : $tan 4 5^{\circ} = \frac{h}{x}$ , so $h = x$ . (1 mark)
(ii) BQ = x + 20. In right $△ P QB$ : $tan 3 0^{\circ} = \frac{h}{x + 20}$ , so $3 h = x + 20$ . (1 mark)
(iii) Substitute $x = h$ : $3 h - h = 20$ , so $h = \frac{20}{3 - 1}$ . (½ mark)
Rationalise: $h = \frac{20 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )} = \frac{20 ( 3 + 1 )}{2} = 10 (3 + 1)$ m. (½ mark)
$h = 10 \times 2.732 = 27.32$ m, and the width $x = h = 27.32$ m. (½ mark)
Check: $B Q = 47.32$ m and $\frac{27.32}{47.32} \approx 0.577 = tan 3 0^{\circ}$ . (½ mark)

Answer: (i) Height PQ = h is shared; $h = x$ (ii) $3 h = x + 20$ (iii) River width = tower height = $10 (3 + 1)$ m, about 27.32 m.

Example 6: Transmission tower on a building · 3 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

Figure: building AB (A top, B foot), tower AC on top with AC = 10 m, so CB = b + 10 where AB = b. P on the ground with PB = d, the shared side. $∠ A P B = 4 5^{\circ}$ , $∠ C P B = 6 0^{\circ}$ . (½ mark)
In right $△ A B P$ : $tan 4 5^{\circ} = \frac{b}{d}$ , so $d = b$ . (½ mark)
In right $△ C B P$ : $tan 6 0^{\circ} = \frac{b + 10}{d}$ , so $b + 10 = 3 d = 3 b$ . (½ mark)
$b (3 - 1) = 10$ , so $b = \frac{10}{3 - 1}$ . (½ mark)
Rationalise: $b = \frac{10 ( 3 + 1 )}{2} = 5 (3 + 1) = 5 \times 2.732 = 13.66$ m, and $d = 13.66$ m. (½ mark)
Check: $C B = 23.66$ m and $\frac{23.66}{13.66} \approx 1.732 = tan 6 0^{\circ}$ . (½ mark)

Answer: Building height $= 5 (3 + 1)$ m, about 13.66 m; P is also $5 (3 + 1)$ m, about 13.66 m, from the building.

Example 7: A car approaching a tower: time to reach the foot · 3 marks · Short answerHarder board level

A car moves at a uniform speed along a straight level road that leads to the foot of a vertical tower. From the car, the angle of elevation of the top of the tower changes from $3 0^{\circ}$ to $4 5^{\circ}$ in 12 seconds. How many more seconds will the car take to reach the foot of the tower? Ignore the height of the observer in the car. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: tower PQ = h (P top, Q foot). The car is at A when the angle is $3 0^{\circ}$ and at B when it is $4 5^{\circ}$ , with A, B, Q in one line. Let the speed be v m/s, so AB = 12v. (½ mark)
In right $△ P QB$ : $tan 4 5^{\circ} = \frac{h}{B Q}$ , so $B Q = h$ . (½ mark)
In right $△ P Q A$ : $tan 3 0^{\circ} = \frac{h}{A Q}$ , so $A Q = 3 h$ . (½ mark)
Same side: $A B = A Q - B Q = h (3 - 1) = 12 v$ , so $\frac{h}{v} = \frac{12}{3 - 1}$ . (½ mark)
Time from B to Q $= \frac{B Q}{v} = \frac{h}{v} = \frac{12 ( 3 + 1 )}{2} = 6 (3 + 1)$ s. (½ mark)
$6 \times 2.732 = 16.39$ s. The speed and height cancel, so the answer does not need either of them. (½ mark)

Answer: The car needs $6 (3 + 1)$ s, about 16.39 s, more to reach the foot of the tower.

Example 8: Guards on opposite sides of a watchtower · 3 marks · Short answerBoard level

A vertical watchtower PQ (P top, Q foot) stands on level ground between two guards A and B, and A, Q, B lie in one straight line. The angle of elevation of P is $3 0^{\circ}$ from A and $6 0^{\circ}$ from B. Guard A is 40 m farther from the foot of the tower than guard B is. Find the height of the tower and the distance between the two guards. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: Q between A and B on one line, PQ = h, $∠ P A Q = 3 0^{\circ}$ , $∠ P B Q = 6 0^{\circ}$ . (½ mark)
In right $△ P Q A$ : $tan 3 0^{\circ} = \frac{h}{A Q}$ , so $A Q = 3 h$ . (½ mark)
In right $△ P QB$ : $tan 6 0^{\circ} = \frac{h}{B Q}$ , so $B Q = \frac{h}{3}$ . (½ mark)
Given $A Q - B Q = 40$ : $3 h - \frac{h}{3} = \frac{2 h}{3} = 40$ , so $h = 203$ m. (½ mark)
$A Q = 3 \times 203 = 60$ m and $B Q = 20$ m. The guards are on opposite sides, so $A B = A Q + B Q = 80$ m. (½ mark)
$h = 20 \times 1.732 = 34.64$ m. Check: $\frac{20 3}{20} = 3$ and $\frac{20 3}{60} = \frac{1}{3}$ . (½ mark)

Answer: Tower height $= 203$ m, about 34.64 m; the guards are 80 m apart.

Example 9: Observer's eye height and a walk · 5 marks · Long answerHarder board level

A boy whose eyes are 1.5 m above level ground walks straight towards a building 31.5 m high at a steady speed. The angle of elevation of the top of the building from his eyes changes from $3 0^{\circ}$ to $6 0^{\circ}$ in 30 seconds. Find his speed and how much longer he will take to reach the building. Use $3 = 1.732$ .

Try it first, then show the solutionSolution

Figure: building PQ (P top, Q foot), PQ = 31.5 m. The boy's eye moves from A to B along a horizontal line 1.5 m above the ground, meeting PQ at R. So RQ = 1.5 m and the vertical side of each right triangle is $P R = 31.5 - 1.5 = 30$ m. (1 mark)
In right $△ P R A$ : $tan 3 0^{\circ} = \frac{30}{A R}$ , so $A R = 303$ m. (1 mark)
In right $△ P R B$ : $tan 6 0^{\circ} = \frac{30}{B R}$ , so $B R = \frac{30}{3} = 103$ m. (1 mark)
Distance walked $A B = 303 - 103 = 203 \approx 34.64$ m in 30 s, so speed $= \frac{20 3}{30} = \frac{2 3}{3} \approx 1.15$ m/s. (1 mark)
Remaining distance $B R = 103$ m, which is half of AB, so it takes half the time: 15 s more. (1 mark)

Answer: Speed $\frac{2 3}{3} \approx 1.15$ m/s; he needs 15 s more. The 1.5 m eye height is subtracted from the building first.

Where students lose marks in Problems with Two Right Triangles

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

Mistake: Writing the farther distance as 60 − x instead of x + 60 on the same side
The student treats the given gap as the total distance from the farther point to the foot, as if the points were on opposite sides.
Wrong working
Same side, AB = 60 m: let BQ = x, so AQ = 60 - x.
Right working
Same side, AB = 60 m: let BQ = x, so AQ = x + 60.
Do this: Draw the figure first. If both points are on one side, the farther distance is the nearer distance plus the gap.
Usually costs: Usually all marks after the figure, since every later equation is wrong.
Mistake: Putting the angle of depression inside the triangle at the top
The depression is measured from the horizontal at the eye, not from the vertical tower.
Wrong working
Depression $3 0^{\circ}$ from a 75 m tower: $tan 3 0^{\circ} = \frac{B Q}{75}$ , so $B Q = 253$ m.
Right working
By alternate angles, the angle at the boat is $3 0^{\circ}$ : $tan 3 0^{\circ} = \frac{75}{B Q}$ , so $B Q = 753$ m.
Do this: Draw the horizontal at the top, mark the depression there, then transfer it to the ground point and write 'alternate angles'.
Usually costs: Most of the marks for that triangle and the final answer.
Mistake: Forgetting the common horizontal in building-and-tower questions
The student uses the full tower height in the elevation triangle instead of only the part above the observer.
Wrong working
Building 12 m, BD = 12 m, elevation $6 0^{\circ}$ : $tan 6 0^{\circ} = \frac{C D}{12}$ , so $C D = 123$ m.
Right working
$tan 6 0^{\circ} = \frac{C E}{12}$ gives $C E = 123$ m, then $C D = C E + E D = 123 + 12$ m.
Do this: Always draw the horizontal through the observer's eye and split the tower into two parts.
Usually costs: 1 to 2 marks for the final height.
Mistake: Rationalising wrongly or rounding too early
Using 1.732 in the denominator before rationalising leads to messy division and rounding errors.
Wrong working
$h = \frac{50}{3 - 1} = \frac{50}{0.73} \approx 68.49$ m.
Right working
$h = \frac{50 ( 3 + 1 )}{( 3 - 1 ) ( 3 + 1 )} = 25 (3 + 1) = 68.3$ m.
Do this: Keep surds until the last line; multiply by the conjugate, then substitute 1.732.
Usually costs: ½ to 1 mark for accuracy.

Exam tips

Draw one clean, labelled figure even if the question does not ask for it; examiners give marks for it in long answers.
Name the triangle with each ratio, such as 'In right triangle PQB', and write the alternate angles reason whenever you transfer a depression.
Leave answers in surd form unless a value of $3$ is given or a decimal is asked for.
In case studies, the 1-mark parts often give the two equations; reuse them exactly in the 2-mark part.
After solving, substitute back into both equations, or check that the bigger angle belongs to the nearer point.

Practice questions on Problems with Two Right Triangles

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

Q1 · 1 mark · MCQBoard level
From two points on level ground on the same side of a vertical tower, in line with its foot, the angles of elevation of the top are $3 0^{\circ}$ and $6 0^{\circ}$ . The points are 24 m apart. The height of the tower is
(a) $123$ m
(b) $243$ m
(c) $83$ m
(d) $12$ m
Hint
Let the nearer distance be x: $h = 3 x$ and $3 h = x + 24$ .
Answer
(a) $123$ m. From $3 x = x + 24$ , $x = 12$ m, so $h = 123$ m.
Q2 · 3 marks · Short answerBoard level
An aeroplane flies horizontally at a constant height of $15003$ m above level ground, moving straight away from an observer on the ground, in the vertical plane through the observer. The angle of elevation of the aeroplane changes from $6 0^{\circ}$ to $3 0^{\circ}$ in 15 seconds. Find the speed of the aeroplane in m/s and in km/h.
Hint
Find the horizontal distance of the plane from the observer at each moment; the difference is the distance flown in 15 s.
Answer
At $6 0^{\circ}$ : distance $= \frac{1500 3}{3} = 1500$ m. At $3 0^{\circ}$ : distance $= 15003 \times 3 = 4500$ m. Distance flown = 3000 m in 15 s, so speed = 200 m/s = $200 \times \frac{18}{5} = 720$ km/h.
Q3 · 3 marks · Short answerHarder board level
A car moves at a uniform speed of 36 km/h along a straight level road towards the foot of a vertical tower. From the car, the angle of elevation of the top of the tower changes from $3 0^{\circ}$ to $6 0^{\circ}$ in 10 seconds. Find the height of the tower, and how many more seconds the car takes to reach its foot. Ignore the observer's height. Use $3 = 1.732$ .
Hint
Convert 36 km/h to m/s first. The distance covered in 10 s is $3 h - \frac{h}{3}$ .
Answer
Speed $= 36 \times \frac{5}{18} = 10$ m/s, so the gap is 100 m. $3 h - \frac{h}{3} = \frac{2 h}{3} = 100$ , so $h = 503 \approx 86.6$ m. Distance left $= \frac{h}{3} = 50$ m, so the car takes 5 s more.
Q4 · 5 marks · Long answerHarder board level
From the top A of a vertical building AB, the angle of elevation of the top C of a taller vertical tower CD is $6 0^{\circ}$ and the angle of depression of its foot D is $3 0^{\circ}$ . Both stand on the same level ground.
(a) Prove that the tower is four times as tall as the building.
(b) If the tower is 40 m tall, find the height of the building and the distance between the building and the tower.
(c) Check your answer by finding the angle of elevation of A as seen from D. Use $3 = 1.732$ .
Hint
Draw AE horizontal meeting CD at E, so ED = AB = b and AE = BD = d. Write d in terms of b from the depression, then CE in terms of b.
Answer
(a) $∠ A D B = 3 0^{\circ}$ (alternate angles), so $tan 3 0^{\circ} = \frac{b}{d}$ gives $d = 3 b$ . In right $△ A E C$ : $C E = d tan 6 0^{\circ} = 3 b \times 3 = 3 b$ . So $C D = C E + E D = 3 b + b = 4 b$ . (b) $4 b = 40$ , so $b = 10$ m and $d = 103 \approx 17.32$ m. (c) $tan ∠ A D B = \frac{10}{10 3} = \frac{1}{3}$ , so the angle is $3 0^{\circ}$ , equal to the angle of depression, as alternate angles require.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: At a school fair, a hot-air balloon is tied to a point Q on level ground and rises vertically above Q at a steady speed of 1.5 m/s. A student stands at a fixed point A on the ground. At one moment the angle of elevation of the balloon from A is $3 0^{\circ}$ ; exactly 40 seconds later it is $6 0^{\circ}$ . Ignore the student's height.
(i) Which length is common to the two right triangles in this situation? (1 mark)
(ii) How far does the balloon rise between the two readings? (1 mark)
(iii) Find the distance of A from Q and the height of the balloon at each reading. Use $3 = 1.732$ . (2 marks)
Hint
The ground distance AQ = d stays fixed. Write both heights in terms of d; their difference is the rise.
Answer
(i) The ground distance AQ. (ii) $1.5 \times 40 = 60$ m. (iii) Heights are $\frac{d}{3}$ and $3 d$ , so $3 d - \frac{d}{3} = \frac{2 d}{3} = 60$ , giving $d = 303 \approx 51.96$ m. Heights: 30 m at the first reading and 90 m at the second (difference 60 m, as required).
Q6 · 3 marks · Short answerHarder board level
A boy whose eyes are 1.5 m above the ground stands on level ground in front of a building 31.5 m high. The angle of elevation of the top of the building from his eyes is $4 5^{\circ}$ . He walks towards the building until the angle is $6 0^{\circ}$ . How far did he walk? Use $3 = 1.732$ .
Hint
The vertical side of both triangles is $31.5 - 1.5$ m.
Answer
Vertical side $= 30$ m. At $4 5^{\circ}$ the distance is $\frac{30}{t a n 4 5 ^{\circ}} = 30$ m; at $6 0^{\circ}$ it is $\frac{30}{3} = 103 \approx 17.32$ m. He walked $30 - 103 \approx 12.68$ m.
Q7 · 5 marks · Long answerHarder board level
A building and a tower stand on the same level ground, and the sum of their heights is 80 m. From the foot of the building, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From the top of the building, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of each and the distance between them. Use $3 = 1.732$ .
Hint
Let the distance be d. Write the tower height and the part of the tower above the building's top in terms of d, then use the sum.
Answer
Let the distance be $d$ . From the foot: tower $T = d tan 6 0^{\circ} = 3 d$ . From the top of the building (horizontal line at the building's height $b$ ): $T - b = d tan 3 0^{\circ} = \frac{d}{3}$ , so $b = 3 d - \frac{d}{3} = \frac{2 d}{3}$ . Sum: $3 d + \frac{2 d}{3} = \frac{5 d}{3} = 80$ , so $d = 163 \approx 27.71$ m. Then $T = 48$ m and $b = 32$ m. Check: $48 + 32 = 80$ and $\frac{48 - 32}{16 3} = \frac{1}{3} = tan 3 0^{\circ}$ .
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose Rahul and Meena are looking at the top of a temple tower on level ground, both in one straight line with its foot. Rahul says: "I am 30 m from the foot and I see the top at $6 0^{\circ}$ ." Meena says: "I see the top at $3 0^{\circ}$ , and I am 60 m from Rahul." Ignore their heights.
(i) Find the height of the tower. (1 mark)
(ii) Find Meena's distance from the foot of the tower. (1 mark)
(iii) Decide whether Meena is on the same side of the tower as Rahul or on the opposite side, with a reason. If she had been on the other side at the same distance from the foot, how far apart would they be? (2 marks)
Hint
Do not assume a layout; use the numbers to decide it.
Answer
(i) $h = 30 tan 6 0^{\circ} = 303 \approx 51.96$ m. (1)
(ii) Meena's distance $= \frac{h}{t a n 3 0 ^{\circ}} = 303 \times 3 = 90$ m. (1)
(iii) On the same side the gap would be $90 - 30 = 60$ m, which matches what Meena says; on opposite sides it would be $90 + 30 = 120$ m. So she is on the same side, farther away (her smaller angle also fits a farther point). On the other side they would be 120 m apart. (2)
Q9 · 1 mark · MCQBoard level
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 $4 5^{\circ}$ and $6 0^{\circ}$ . The distance between the points is
(a) $h (1 + \frac{1}{3})$
(b) $h (1 - \frac{1}{3})$
(c) $h (3 + 1)$
(d) $2 h$
Hint
Opposite sides: add the two distances from the foot.
Answer
(a). The distances are $h cot 4 5^{\circ} = h$ and $h cot 6 0^{\circ} = \frac{h}{3}$ ; on opposite sides they add.
Q10 · 1 mark · Assertion-reasonHarder board level
Assertion (A): For two points on level ground on the same side of a tower, in line with its foot, at which the angles of elevation of the top are $3 0^{\circ}$ and $6 0^{\circ}$ , the farther point is exactly three times as far from the foot as the nearer point.
Reason (R): As a point on level ground moves closer to the foot of a tower, the angle of elevation of the top increases.
(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
Compute both distances in terms of h.
Answer
(b). Distances are $3 h$ and $\frac{h}{3}$ , and $3 h \div \frac{h}{3} = 3$ , so A is true. R is true, but it only tells you which point is nearer; the factor 3 comes from $\frac{t a n 6 0 ^{\circ}}{t a n 3 0 ^{\circ}} = 3$ , so R does not explain A.
Q11 · 2 marks · Very short answerBoard level
From a point 20 m from the foot of a vertical tower on level ground, the angle of elevation of the top is $6 0^{\circ}$ . How far from the foot, on the same line, is the point from which the angle of elevation is $3 0^{\circ}$ ?
Hint
The height is the side the two triangles share.
Answer
Height $= 20 tan 6 0^{\circ} = 203$ m (1 mark). Distance at $3 0^{\circ}$ $= \frac{20 3}{t a n 3 0 ^{\circ}} = 203 \times 3 = 60$ m (1 mark).

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.Same side, nearer distance x, gap d: what is the farther distance?Show answer
$x + d$
2.Opposite sides: how do the two horizontal distances combine?Show answer
They add to give the total distance.
3.Same side, $3 0^{\circ}$ and $6 0^{\circ}$ , points d apart: height?Show answer
$\frac{3}{2} d$
4.Rationalise $\frac{1}{3 - 1}$ .Show answer
$\frac{3 + 1}{2}$
5.Which point has the larger angle of elevation?Show answer
The one nearer to the tower.
6.From a building of height b, depression of the tower's foot is $4 5^{\circ}$ . Distance between them?Show answer
b

Problems with Two Right Triangles: frequently asked questions

How do I know which side the two triangles share?

Look for the length that belongs to both triangles in your figure. When one tower is seen from two ground points, the tower's height is shared. When one observer looks up and down, the horizontal distance is shared.

Can a board question have three right triangles?

No. The 2026-27 syllabus says problems should not involve more than two right triangles, and the angles are limited to 30, 45 and 60 degrees.

Should I give the answer in surd form or as a decimal?

Give the exact surd form unless the question gives a value such as root 3 = 1.732 or asks for a decimal. If a value is given, show the surd form first, then substitute.

Do I need to draw the figure if marks are not mentioned for it?

Yes. Examiners usually give a mark for a correct figure in long answers, and the figure tells you whether to add or subtract distances.

Is a formula like h = d divided by (cot A minus cot B) accepted?

It is safer to write both tan equations and solve them, because step marks are given for each equation. Use the formula only to check your final answer.