Exercise 12.2 — Two Triangle Problems
Problems based on figures with two triangles.
Exercise 12.2 — Applications of Trigonometry (Heights and Distances)
Class 10 Mathematics · CBSE, Telangana & Andhra Pradesh Syllabus · Chapter 12
Exercise 12.2 is the main problem-solving exercise of Chapter 12, Applications of Trigonometry, in Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). Every question in this exercise is a real-world "heights and distances" puzzle — towers, temples, statues, wires, poles, buildings, and even a jet plane — that is solved using nothing more than right-angle triangles and the three trigonometric ratios sin, cos, and tan at the standard angles 30°, 45°, and 60°.
Below you will find a complete, step-by-step explanation of all 10 problems in this exercise, along with the underlying concepts of angle of elevation and angle of depression, a quick-reference trigonometric ratio table, fully labelled diagrams for every problem, and a summary table of every final answer — everything you need to revise this exercise thoroughly for your board exams.
Angle of Elevation and Angle of Depression
Both terms describe the angle between a person's horizontal line of sight and the line joining their eye to a distant object. The angle of elevation is used when looking upward at something above the horizontal — the top of a tower, a temple, or a flying jet. The angle of depression is used when looking downward at something below the horizontal — typically used when an observer is standing at a height, such as on top of a building, looking down at the foot of another object.
Left: looking up at a tower from the ground (angle of elevation). Right: looking down from a height at the foot of an object (angle of depression).
Trigonometric Ratios You Will Need
Almost every problem in this exercise only ever uses the angles 30°, 45°, and 60°. Keeping these standard values memorised (rather than looking them up each time) will save valuable time in the exam:
| Ratio | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan θ | 0 | 1/√3 | 1 | √3 | Not defined |
tan θ = Opposite side / Adjacent side | sin θ = Opposite side / Hypotenuse
- Use tan θ whenever you know (or want to find) the height of a vertical object and its horizontal distance from the observer.
- Use sin θ whenever the length of a slanting line — like a wire or a rope — is involved, since the hypotenuse is what connects the slant length to the height.
- Throughout this exercise, take √3 = 1.732 wherever a decimal value is required, as instructed in the textbook.
Problem 1 — Height of a TV Tower and Width of a Road
Question: A TV tower stands on one side of a road. From a point directly opposite the tower, the angle of elevation of its top is 60°. From a second point, 10 m further back along the same line, the angle of elevation is 30°. Find the height of the tower and the width of the road.
C is directly opposite the tower (60° elevation); D is 10 m further back (30° elevation). BC = x is the road width, AB = h is the tower height.
Let the tower height AB = h and the road width BC = x. Applying tan to both right triangles ABC and ABD:
Problem 2 — Distance Walked Towards a Temple
Question: A 1.5 m tall boy looks at the top of a 30 m temple. As he walks towards it, the angle of elevation of the top increases from 30° to 60°. Find the distance he walked.
C and F are the boy's two positions; the dashed line is his eye level, 1.5 m above the ground, leaving AE = 30 − 1.5 = 28.5 m above eye level.
Since the boy's eyes are 1.5 m above the ground, the part of the temple above his eye level is AE = 30 − 1.5 = 28.5 m. Let EF = the distance from the nearer point F to directly below A, and FC = the distance walked.
Problem 3 — Height of a Statue on a Pedestal
Question: A statue stands on a 2 m tall pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60°, and of the top of the pedestal is 45°. Find the height of the statue.
BC = 2 m is the pedestal, AB = h is the statue. From D, the pedestal top C is seen at 45° and the statue top A at 60°.
Let the statue height AB = h and let CD be the horizontal distance from the observation point D to the pedestal.
Problem 4 — Height of a Cell Tower Seen From a Building
Question: From the top of a building, the angle of elevation of the top of a cell tower is 60°, and the angle of depression of its foot is 45°. If the building is 7 m from the tower, find the height of the tower.
A is the eye-level point at the top of the building. C is the top of the cell tower (60° elevation); D is its foot (45° depression).
Since the eye-level line at the top of the building is parallel to the ground, the angle of depression to D equals the angle ADE = 45° on the other side. The horizontal distance AE = BD = 7 m.
Problem 5 — Length of Wire Cut From an Electric Pole
Question: An 18 m wire was tied to a pole at an angle of elevation of 30° with the ground. It was then cut and re-tied at an angle of elevation of 60°. Find the length of wire that was cut off.
AD = 18 m is the original wire (30° with the ground). After cutting, the shorter wire AC is re-tied at 60°.
Let the pole height AB = h. The original wire AD = 18 m is the hypotenuse of the 30° triangle, and the new wire AC is the hypotenuse of the 60° triangle — both use the sine ratio since they are slanting lengths.
Problem 6 — Height of a Building Using a Tower's Height
Question: The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 30 m high, find the height of the building.
AB is the building, CD = 30 m is the tower. The crossing lines of sight create two right triangles sharing the base BD.
Let AB = building height, BD = distance between the two feet. Use the tower's triangle first to find BD, then the building's triangle to find AB.
Problem 7 — Two Equal Poles on Either Side of a Road
Question: Two poles of equal height stand on either side of a 120-feet-wide road. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the distances of the point from each pole.
E is the point of observation between the two equal poles, BD = 120 ft is the full road width.
Let BE = x and ED = 120 − x. Since both poles have the same height, their two expressions for height must be equal.
Problem 8 — Tower Height Using Complementary Angles
Question: From two points 4 m and 9 m away from a tower (in a straight line with its foot), the angles of elevation of the top are complementary (they add up to 90°). Find the height of the tower.
C is 4 m from the tower with elevation θ; D is 9 m away with elevation (90° − θ), since the two angles are complementary.
Let the tower height AB = h, with ∠ACB = θ so that ∠ADB = 90° − θ. Writing tan for both triangles and multiplying them together eliminates θ completely.
Problem 9 — Speed of a Jet Plane
Question: The angle of elevation of a jet flying at a constant height of 1500√3 m is 60° from a point A. After 15 seconds, it changes to 30°. Find the speed of the jet.
The plane flies horizontally at a constant height of 1500√3 m, from position B (60° elevation) to D (30° elevation), 15 seconds later.
Both BC and DE equal the constant flying height, 1500√3 m. Find the horizontal distances AC and AE, then subtract to get the distance flown.
Problem 10 — Ratio of Heights of a Tower and a Building
Question: The angle of elevation of the top of a tower from the foot of a building is 30°, and the angle of elevation of the top of the building from the foot of the tower is 60°. Find the ratio of the heights of the tower and the building.
AB = building, CD = tower, sharing the common base BD. The tower's top is seen at 30° from B, and the building's top at 60° from D.
Let AB = building height, CD = tower height, and BD = the common base distance between them.
Common Mistakes to Avoid
- Using tan instead of sin for slanting lengths: In wire/rope problems (like Problem 5), the given length is the hypotenuse, not a side — always use sin θ = opposite/hypotenuse, not tan θ.
- Forgetting to add or subtract the observer's height: In Problem 2, the boy's eye-level height (1.5 m) must be subtracted from the temple's total height before applying the trig ratios.
- Mixing up elevation and depression: The angle of depression from the top is always equal to the angle of elevation seen from the bottom, since the eye-level line is parallel to the ground — confusing the two often leads to placing the angle on the wrong side of the triangle.
- Not simplifying surds correctly: Expressions like 1/√3 should be rationalised to √3/3 before being multiplied or compared, to avoid arithmetic slips.
- Skipping the diagram: Every heights-and-distances question becomes far easier once a clearly labelled diagram is drawn first — attempting the algebra without a diagram is the single biggest cause of errors in this exercise.
- Forgetting the units conversion at the end: Problem 9 asks for speed in km/hr, but the natural working gives m/s — always check what unit the final answer is required in.
Quick Reference — All 10 Answers at a Glance
| Q.No | Scenario | Key Result |
|---|---|---|
| 1 | TV tower & road width | Tower = 5√3 m ≈ 8.66 m; Road = 5 m |
| 2 | Boy walking towards a temple | Distance walked = 19√3 m ≈ 32.91 m |
| 3 | Statue on a pedestal | Statue height = 2(√3−1) m ≈ 1.46 m |
| 4 | Cell tower from a building | Tower height = 7(√3+1) m ≈ 19.12 m |
| 5 | Wire cut from a pole | Length cut = 18 − 6√3 m ≈ 7.61 m |
| 6 | Building height from tower | Building = 10 m |
| 7 | Two equal poles on a road | Pole height = 30√3 ft ≈ 51.96 ft; 30 ft & 90 ft |
| 8 | Complementary angles tower | Tower height = 6 m |
| 9 | Speed of a jet plane | Speed = 200 m/s = 720 km/hr |
| 10 | Tower-to-building ratio | Ratio = 1 : 3 |
What This Exercise Prepares You For
Exercise 12.2 builds directly on the trigonometric ratios and identities introduced earlier in the chapter, so a quick revision of the basics of trigonometric ratios is useful before attempting these problems, especially the values of sin, cos, and tan at 30°, 45°, and 60°.
The right-triangle reasoning practised here also reinforces the properties of similar triangles from the Triangles chapter, since every heights-and-distances figure is essentially built from one or more right triangles sharing a common side. Once this exercise feels comfortable, it is also worth revisiting surds and radical simplification, since answers in this exercise are almost always left in terms of √3.