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Lesson 2 of 2

Some Applications of Trigonometry · Lesson 2 of 2

Chapter Summary and Practice

“Turn angles of elevation and depression into practical ways to measure the world around you.”

Learning Objectives

• By the end of this lesson, you should be able to. • Recall the key language and method of height-and-distance problems. • Decide quickly which trigonometric ratio to use. • Solve mixed problems involving elevation, depression, observer height and multiple triangles. • Avoid the common diagram and interpretation mistakes that cost marks.

This chapter is less about memorising new formulas and more about modelling a situation correctly. The core skill is to draw the right triangle, identify the known angle and sides, and then choose the ratio that connects what you know to what you need.

Line of Sight

The line of sight is the straight line from the observer's eye to the point being viewed. In diagrams, it usually becomes the hypotenuse of the right triangle.

Angle of Elevation

When the object is above the observer's horizontal eye level, the angle between the horizontal and the upward line of sight is the angle of elevation.

Angle of Depression

When the object is below the observer's horizontal eye level, the angle between the horizontal and the downward line of sight is the angle of depression. Because horizontal lines are parallel, this angle can usually be transferred to an equal angle of elevation inside the right triangle.

How to Choose the Right Ratio

Sides involvedUseful ratio
opposite + adjacenttan θ = opposite/adjacent
opposite + hypotenusesin θ = opposite/hypotenuse
adjacent + hypotenusecos θ = adjacent/hypotenuse

Problem-Solving Checklist

Before Writing Any Equation

• Is the object vertical and the ground horizontal? • Where is the observer's eye? • Is the given angle elevation or depression? • Which segment is the actual horizontal distance? • Is there one right triangle or more than one? • Does the observer's height need to be added at the end?

Common Mistakes

Avoid These

• Treating a line of sight as horizontal distance. • Forgetting the eye height. • Using an angle of depression at the wrong vertex. • Mixing lengths from two different triangles. • Applying a trig ratio before deciding which angle is the reference angle.

Guided Practice

Worked Example: Rope and Pole

Problem
A 24 m rope is stretched from the top of a vertical pole to the ground and makes an angle of 30° with the ground. Find the pole height.

  1. 1.The rope is the hypotenuse because it is opposite the right angle.
  2. 2.The pole height is opposite the 30° angle.
  3. 3.Use sine: sin 30° = height/24.
  4. 4.1/2 = height/24.
  5. 5.Height = 12 m.
Worked Example: Observer Walks Towards a Building

Problem
A 1.5 m tall observer looks at the top of a 25.5 m building. The angle of elevation changes from 30° to 60° after the observer walks towards the building. Find the distance walked.

  1. 1.The vertical height above the observer's eyes is 25.5 - 1.5 = 24 m.
  2. 2.Let the first horizontal distance be x.
  3. 3.tan 30° = 24/x, so x = 24√3.
  4. 4.Let the second horizontal distance be y.
  5. 5.tan 60° = 24/y, so y = 24/√3 = 8√3.
  6. 6.Distance walked = x - y = 24√3 - 8√3 = 16√3 m.
Worked Example: Tower on Top of a Building

Problem
From a point on the ground, the angle of elevation of the top of an 18 m building is 45°, while the angle of elevation of the top of a tower fixed on the building is 60°. Find the tower height.

  1. 1.Let the horizontal distance from the observation point to the building be d.
  2. 2.From the 45° triangle: tan 45° = 18/d, so d = 18 m.
  3. 3.Let the tower height be h. Total height is 18 + h.
  4. 4.From the 60° triangle: tan 60° = (18 + h)/18.
  5. 5.√3 = (18 + h)/18.
  6. 6.18 + h = 18√3.
  7. 7.h = 18(√3 - 1) m.
Worked Example: Equal Poles Across a Road

Problem
Two equal poles stand on opposite sides of a 60 m wide road. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of each pole.

  1. 1.Let the distance from the point to the nearer pole be x m. Then the distance to the other pole is 60 - x.
  2. 2.Let the common pole height be h.
  3. 3.For the 60° pole: tan 60° = h/x, so h = x√3.
  4. 4.For the 30° pole: tan 30° = h/(60 - x), so h = (60 - x)/√3.
  5. 5.Equate the two expressions for h:
  6. 6.x√3 = (60 - x)/√3.
  7. 7.3x = 60 - x.
  8. 8.4x = 60, so x = 15.
  9. 9.Therefore h = 15√3 m.
Worked Example: Ships Seen from a Lighthouse

Problem
From the top of a 75 m lighthouse, the angles of depression of two ships on the same side are 45° and 30°. Find the distance between the ships.

  1. 1.The angles of depression equal the corresponding angles of elevation from the ships.
  2. 2.Let the nearer ship be at distance x from the lighthouse foot.
  3. 3.tan 45° = 75/x, so x = 75 m.
  4. 4.Let the farther ship be at distance y.
  5. 5.tan 30° = 75/y, so y = 75√3 m.
  6. 6.Distance between ships = y - x = 75(√3 - 1) m.

Quiz

Quick check

An object is above the observer's horizontal level. Which angle is formed?

Quick check

If vertical height and horizontal distance are the two sides involved, which ratio is usually most direct?

Quick check

A 1.6 m tall observer calculates 20 m as the height above eye level. What is the full object height?

Quick check

Why can an angle of depression often be used as an equal angle of elevation?

Quick check

A tower casts a shorter shadow when the Sun's altitude is:

Before You Finish the Chapter

Ask yourself: • Can I draw a correct diagram from a word problem? • Can I place an elevation or depression angle at the correct vertex? • Can I choose sin, cos or tan without trial and error? • Can I handle observer height separately? • Can I solve two-triangle problems by defining shared distances clearly?

Practice Problems

Practice Questions
  1. A 22 m rope is tied from the top of a vertical pole to the ground and makes an angle of 30° with the ground. Find the height of the pole.
  2. A tree breaks and its top touches the ground 12 m from the foot, making an angle of 30° with the ground. Find the original height of the tree.
  3. A slide is 2 m high and inclined at 30° to the ground. Another slide is 4 m high and inclined at 60°. Find the length of each slide.
  4. The angle of elevation of the top of a tower from a point 36 m away is 30°. Find the tower height.
  5. A kite flies 72 m above the ground. Its taut string makes an angle of 60° with the ground. Find the string length.
  6. A 1.5 m tall boy observes the top of a 28.5 m building. The angle of elevation changes from 30° to 60° as he walks towards it. Find how far he walks.
  7. From a point on the ground, the angles of elevation of the bottom and top of a tower mounted on an 18 m building are 45° and 60°. Find the tower height.
  8. A 1.8 m statue stands on a pedestal. From the same point on the ground, the angles of elevation of the top of the statue and pedestal are 60° and 45°. Find the pedestal height.
  9. The angle of elevation of the top of a building from the foot of a 48 m tower is 30°, while the angle of elevation of the top of the tower from the foot of the building is 60°. Find the building height.
  10. Two equal poles stand on opposite sides of an 84 m road. From a point between them, the angles of elevation are 60° and 30°. Find the pole height and the distances from the point to the poles.
  11. A TV tower stands on one bank of a canal. From a point directly opposite on the other bank, the angle of elevation is 60°. From another point 18 m farther away along the same line, the angle is 30°. Find the tower height and canal width.
  12. From the top of a 9 m building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the tower height.
  13. From the top of a 90 m lighthouse, the angles of depression of two ships on the same side are 30° and 45°. Find the distance between the ships.
  14. A 1.3 m tall girl sees a balloon moving horizontally at 91.3 m above ground. Its angle of elevation changes from 60° to 30°. Find how far it travels.
  15. A car approaches the foot of a tower at uniform speed. From the top of the tower, its angle of depression is 30°. Six seconds later the angle becomes 60°. Find the time the car takes from the second position to reach the tower.

Key Takeaways

Key Takeaways

• Heights and distances problems are modelling problems first and trigonometry problems second. • A good diagram usually reveals the correct ratio immediately. • Elevation means looking up; depression means looking down. • Tangent is the most common ratio because many problems involve vertical height and horizontal distance. • Complex-looking questions often reduce to two simple right triangles connected by a shared side.

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Some Applications of Trigonometry

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