Chapter 9, Some Applications of Trigonometry, is where Class 10 trigonometry first meets the real world. Every question turns a real object — a tower, a pole, a kite, a building — into a right triangle, labels one angle, and asks for a missing side.
This page routes you to the right question and tells you which method it wants, before you open the solutions.
Chapter 9 has exactly one exercise: Exercise 9.1, with 15 heights-and-distances questions. Below: the chapter’s vocabulary, a question-by-question routing table grouped by method, the two techniques that run through every problem, the mistakes that cost marks, and a link to the full step-by-step solutions.
Chapter 9 at a Glance: Heights and Distances
The chapter’s job in one sentence: identify the right triangle hidden in the situation, then find one missing side. Three terms carry every question, so fix them before anything else.
The line of sight is the straight line drawn from the observer’s eye to the point being viewed.
The angle of elevation is the angle the line of sight makes with the horizontal when you raise your head to look at the point — the point is above the horizontal level.
The angle of depression is that same angle when you lower your head to look at a point below the horizontal level.
One finishing step runs through the whole chapter: when the observer stands on the ground beside the object, the total height is the triangle’s height plus the observer’s own height. Example 3 shows the pattern — it adds the observer’s 1.5 m to the height the triangle gives.
The ratios themselves come from Chapter 8, Introduction to Trigonometry; this chapter only applies them.
Exercise 9.1 Directory: All 15 Questions, Grouped by Method
All 15 questions sit in one exercise, and they fall into six method clusters. Find your question below, read what its cluster tests, and you know the ratio before you draw a single line.
- Q1, Q3, Q5 — the sloping side is the hypotenuse. Q1 is the circus rope tied from the pole top to the ground at 30°; Q3 the two play-park slides, 1.5 m and 3 m high; Q5 the kite string at 60 m. In each, the rope, slide or string is the hypotenuse paired with a known vertical — use sine.
- Q2, Q4 — one triangle, one tangent. Q2 is the storm-broken tree whose top touches the ground 8 m from its foot; Q4 the plain tower seen from 30 m away. Both are a single right triangle with the ground distance known and one height wanted — solve with tangent (or cotangent).
- Q6, Q14 — two angles, two positions, subtract. Q6 is the 1.5 m boy watching a 30 m building as the elevation rises from 30° to 60° while he walks; Q14 the 1.2 m girl watching a balloon at 88.2 m as its elevation falls from 60° to 30°. Each writes two tangent equations, gets two horizontal distances and subtracts — and both hide an eye-height step.
- Q7, Q8, Q9, Q10 — stacked heights on a shared base. The transmission tower on a 20 m building, the 1.6 m statue on a pedestal, a building and a tower looking at each other, and two equal poles across an 80 m road. Two triangles share one base, tangent at two angles, then subtract the pair to isolate the height the question names.
- Q11 — the canal TV tower, shadow style. Elevation 60° from the opposite bank, 30° from a point 20 m back. Two tangent equations share the tower’s height with a 20 m gap between the bases — solve them together for the height and the canal’s width, exactly as Example 5’s growing shadow does.
- Q12, Q13, Q15 — depression angles that bend inwards. A cable tower seen from a 7 m building, two ships from a 75 m lighthouse, a car tracked as its depression angle falls from 30° to 60° in six seconds. Each carries a depression angle that must be carried into the triangle by the alternate angles rule before any tangent works.
Four questions carry textbook figures: Q1 and Q7 share Fig 9.11 (the rope tied from the pole top to the ground), Q11 uses Fig 9.12 (the tower across the canal with both ground points marked) and Q14 uses Fig 9.13 (the balloon at two points on its horizontal path). The step-by-step working for all 15 questions lives on the Exercise 9.1 solutions page.
The Two Methods That Run Through the Chapter
Every problem in this chapter is one of two shapes. Here they are.
Method 1: One Right Triangle, One Ratio
Pick the ratio by which sides feature in the problem — you know some, you want one.
- Tangent when the opposite and adjacent sides are the pair — \( \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \). Example 1 uses it with the 15 m ground distance.
- Sine when the hypotenuse is in play — \( \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} \). Example 2 uses it to find the ladder’s length.
- Cotangent when the adjacent side is the unknown — \( \cot \theta = \frac{\text{adjacent}}{\text{opposite}} \). Example 2 uses it for the ladder’s distance from the pole.
The rule of thumb: name the sides you know and the side you want, then take the ratio whose fraction contains exactly that pair.
Method 2: Two Triangles, Two Tangent Equations, One Shared Height
Two reference points give two angles on the same object, and both right triangles share the same vertical side. Write one tangent equation per triangle and solve the pair together. Example 4 does this for a flagstaff on a building; Example 5 does it for a shadow that grows 40 m as the sun drops from 60° to 30°.
The Supporting Trick: Bending a Depression Angle Inwards
An angle of depression is measured outside the triangle, between the line of sight and the horizontal. The two horizontal lines are parallel and the line of sight is a transversal, so the alternate angles rule carries the depression angle into the triangle unchanged — that equal angle is the one you use.
Example 6 works this way for the 8 m building, and Q13 and Q15 depend on it.
Common Mistakes in Heights and Distances Questions
- Forgetting the observer’s eye height. When the observer stands on the ground beside the object, the triangle’s height is one part and the observer’s height is the other. Example 3 adds 1.5 m; Q6 hides a 1.5 m boy and Q14 a 1.2 m girl. Skip the add and the answer comes out short exactly by that amount.
- Using a depression angle as if it sat inside the triangle. It does not — it sits outside, between the line of sight and the horizontal. Carry it in through the alternate angles rule first, as Q13 and Q15 require.
- Reaching for cos when the side opposite the angle is the unknown. Cosine pairs the adjacent side with the hypotenuse. If the unknown is opposite the given angle, sine or tangent is the ratio — cos guarantees a wrong fraction.
- Stopping at the intermediate height. Q7 asks for the transmission tower, not the 20 m building beneath it; Example 4 wants the flagstaff, not the building. Solve the taller triangle, solve the shorter, then subtract to give the height the question actually names.
- Leaving a surd in the denominator. A fraction like \( \frac{8}{\sqrt{3}-1} \) should be rationalised to \( 4(\sqrt{3}+1) \) by multiplying top and bottom by \( \sqrt{3}+1 \), exactly as the textbook does.
- Substituting \( \sqrt{3} \approx 1.73 \) too early. The surd is exact; a rounded decimal introduces error that compounds through the next line. The textbook substitutes the decimal only when the question says so, and keeps the surd to the final step otherwise.
How to Use This Page Before You Solve
- Find your question number in the Exercise 9.1 directory above and read its cluster.
- Name the ratio the cluster calls for — sine, tangent or cotangent.
- Draw the right triangle first, labelling the horizontal, the vertical and the line of sight; every worked example in the chapter draws before it solves.
- Write the single tangent or sine equation once, solve for the missing side, then do the final add-or-subtract step the question actually asks for.
- Open the Exercise 9.1 solutions page for the full working, and check your answer against it.
For the rest of the course, browse the Class 10 Maths hub, more Class 10 notes, or the full notes index. Next up in the sequence is Chapter 10, Circles. You can verify every figure and episode in this chapter against the official NCERT Class 10 Mathematics textbook.
| Exercise | What you get |
|---|---|
| Exercise 9.1 | Solved |
FAQs on Some Applications of Trigonometry
Why does Class 10 Maths Chapter 9 have only one exercise?
Because all 15 questions of the chapter sit together in Exercise 9.1. There is no Exercise 9.2 or 9.3 — the whole of heights and distances is this one set, grouped within it by method.
How do I decide whether to use sin, tan or cot in an application question?
Name the sides you know and the side you want, then pick the ratio whose fraction contains exactly that pair. Sine pairs the opposite side with the hypotenuse, tangent pairs opposite with adjacent, and cotangent swaps tangent, so it is handy when the adjacent side is the unknown, as in Example 2.
What is the difference between an angle of elevation and an angle of depression?
For elevation, the point being viewed is above the horizontal and you raise your head — the angle sits above the horizontal line. For depression, the point is below and you lower your head — the angle sits below the horizontal. Both are measured from the horizontal, never from the vertical.
When do I have to add the observer’s height to my answer?
Whenever the observer stands on the ground next to the measured object, the total height is the triangle’s part plus the observer’s eye height, because the horizontal runs from the observer’s eye, not from ground level. Example 3 adds 1.5 m, and Q6 and Q14 hide the same step with a 1.5 m boy and a 1.2 m girl.
Should my final answer be 15√3 or a decimal like 25.95?
The surd form, \( 15\sqrt{3} \) for instance, is exact, so keep it unless the question asks for a decimal. A value like 25.95 is the rounded version and only appears when the question tells you to take \( \sqrt{3} = 1.73 \); the textbook makes that substitution explicitly in Examples 2 and 4.
Reference: NCERT textbooks (CBSE).