Exercise 12.1 — Single Triangle Problems
Problems based on figures with one triangle.
Exercise 12.1 — Applications of Trigonometry
Exercise 12.1 is from Chapter 12, Applications of Trigonometry, of Class 10 Mathematics — prescribed by the CBSE, Telangana SSC, and Andhra Pradesh SSC syllabi. This exercise applies trigonometric ratios (sin, cos, tan) to real-life situations involving heights, distances, angles of elevation, and angles of depression.
Unlike Chapter 11 (Introduction to Trigonometry) where you learnt the ratios, this chapter is about using them to solve practical problems — calculating how tall a tower is, how long a ladder should be, or how wide a river is, based on angles measured from the ground or from a height.
Key Concepts: Angle of Elevation vs Angle of Depression
- Angle of Elevation — The angle measured above the horizontal when an observer looks up at an object (e.g., top of a tower).
- Angle of Depression — The angle measured below the horizontal when an observer looks down at an object. By alternate angles property, the angle of depression equals the angle of elevation from the target point.
- The right angle is always at the foot of the vertical object (tower, pole, tree). The horizontal distance is the base, and the height is the opposite side.
Trigonometric Values You Must Memorise
All 10 problems in Exercise 12.1 use only the standard angles: 30°, 45°, and 60°. Memorise these cold — you will use them in every single problem.
| Angle (θ) | sin θ | cos θ | tan θ |
|---|---|---|---|
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | 1/√2 | 1/√2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
Exercise 12.1 — All 10 Problems Solved Step by Step
Setting Up the Triangle
Let A = observation point, BC = height of tower, AC = 15 m (horizontal distance). The angle ∠BAC = 45°.
tan A = opposite/adjacent = BC/ACUnderstanding the Setup
D is the break point, B is the foot of the tree, C is where the broken top touches the ground. The broken part DC = AD (same length before and after). ∠DCB = 30°, BC = 6 m.
Step 1 — Find DB (lower part of tree)
Step 2 — Find DC (broken/fallen part = AD)
Step 3 — Total Height = DB + DC
AB = height of slide = 2 m. AC = length of slide (hypotenuse). ∠ACB = 30°.
sin C = opposite/hypotenuse = AB/ACAB = pole height = 15 m. BC = shadow length = 15√3 m. θ = angle of elevation of sun.
AB = pole height = 10 m. ∠BAC = 30° (angle with the pole, at the top). AC = length of rope.
cos A = adjacent/hypotenuse = AB/ACAB = 6 m (building height). ∠DAC = 60° (depression from horizontal). By alternate angles, ∠ACB = 60°. AC = distance from shooter to target (hypotenuse of triangle ABC).
Step 1 — Find BD (effective height)
Step 2 — Length of Ladder (CD)
Step 3 — Distance BC (foot of ladder to foot of pole)
Draw CD parallel to AB (width). ∠DAC = 60° (angle with bank = angle with horizontal). sin A = CD/AC = AB/AC.
AB = observer height = ED = 1.8 m. AE = BD = 13.2 m. CE = height of tree above observer's eye level. CD = total height of palm tree = CE + ED.
Draw perpendicular BD from B to AC. BD is the height for base AC. Find BD first using sin in triangle ABD.
Step 1 — Find Height BD
Step 2 — Area of Triangle ABC
Area = (1/2) × base × height = (1/2) × AC × BDQuick Summary — All 10 Answers
| Q. | Problem Type | Ratio Used | Angle | Answer |
|---|---|---|---|---|
| 1 | Tower height | tan | 45° | 15 m |
| 2 | Broken tree height | tan + cos | 30° | 6√3 m |
| 3 | Length of slide | sin | 30° | 4 m |
| 4 | Angle of sun elevation | tan (reverse) | — | 30° |
| 5 | Length of ropes (3) | cos | 30° | 34.64 m (total) |
| 6 | Arrow distance (depression) | sin | 60° | 4√3 m |
| 7 | Ladder length + ground distance | sin + tan | 60° | 4.8√3 m; 4.16 m |
| 8 | Width of river | sin | 60° | 300√3 m |
| 9 | Palm tree height (observer included) | tan | 45° | 15 m |
| 10 | Area of triangle | sin | 30° | 7.5 cm² |
Common Mistakes to Avoid in Board Exams
- Wrong ratio choice — Always identify which two sides are known/unknown (opposite, adjacent, hypotenuse) before picking sin/cos/tan. Drawing the triangle label prevents this.
- Forgetting to add observer's height — In Problem 9 type questions, the angle is from the observer's eyes, not from the ground. Always add the observer's height to get the full height of the distant object.
- Angle of depression ≠ angle at top of building — The angle of depression is measured from the horizontal at the top, and it equals the angle of elevation at the base only due to alternate angles. Draw the horizontal line explicitly.
- Rationalising the denominator — When your answer is 12/√3, rationalise it: multiply numerator and denominator by √3 to get 12√3/3 = 4√3. Board examiners expect rationalised answers.
- Broken tree confusion — Remember: the fallen part (DC) is equal to the original upper part (AD), because the tree doesn't lose length when it falls. Height = lower part + fallen part.
• Know opposite and adjacent, want angle → use tan
• Know hypotenuse and opposite → use sin
• Know hypotenuse and adjacent → use cos
• Angle is at the top of the pole (with pole as adjacent) → use cos
Board Exam Tips & What This Lesson Prepares You For
Applications of Trigonometry is one of the most consistently tested topics in Telangana SSC, AP SSC, and CBSE Class 10 Board Exams. Problems from Exercise 12.1 appear as 4-mark or 5-mark questions. The broken tree problem, the river width problem, and observer height problems are the most frequently asked.
- Sketch the diagram first — even a rough one — before writing a single equation. 90% of errors happen when students skip the diagram.
- Label every side of the triangle: mark the right angle (90°), the known angle, the known length, and the unknown with "?".
- Write the ratio formula before substituting numbers. Examiners award process marks.
- Check: does your answer make physical sense? A tree cannot be shorter than its shadow if the sun angle is less than 45°.