Key Concepts
- 1Angle of elevation
- 2Angle of depression
- 3Angle of depression = Angle of elevation
- 4Height & Distance strategy
- 5Which trig ratio to use?
- 6Angle of depression
- 7Height & Distance strategy
- 8What is the angle of depression?
- 9A tower is 30 m high. Find the angle of elevation of its top from a point 30 m from its base.
- 10What is the angle of elevation?
Important Formulas & Facts
Angle between horizontal and line of sight when looking UP.
Angle between horizontal and line of sight when looking DOWN.
Alternate interior angles with the horizontal.
1) Draw figure 2) Mark angles 3) Identify right triangle 4) Use trig ratio
Known side + unknown side → pick ratio connecting them. e.g., P and H → sin; B and P → tan
Angle between horizontal and line of sight when looking DOWN.
1) Draw figure 2) Mark angles 3) Identify right triangle 4) Use trig ratio
The angle of depression is the angle formed between the horizontal line of sight and the line of sight downward to an object below the observer.
tan(theta) = opposite/adjacent = 30/30 = 1. theta = 45 degrees.
The angle of elevation is the angle formed between the horizontal line of sight and the line of sight upward to an object above the observer.
Must-Know Questions
Q1A boy finds a bird at 100 m distance at elevation 30°. A girl on a 20 m building on the opposite side finds elevation 45° to the same bird. Find distance of bird from the girl.
Bird height from ground: sin30° = h/100 → h = 50 m. Height of bird above building = 50 − 20 = 30 m. For the girl: sin45° = 30/d → d = 30/sin45° = 30/(1/√2) = 30√2 m ≈ 42.43 m.
Q2A man on a ship 10 m above water observes the angle of elevation of the top of a hill as 60° and angle of depression of the base as 30°. Find the height of the hill.
Let distance from ship to hill base = d. tan30° = 10/d → d = 10√3 m. Let hill height above ship level = H. tan60° = H/d → H = d·tan60° = 10√3 × √3 = 30 m. Total height of hill = H + 10 = 30 + 10 = 40 m.
Q3A tower is 20 m high surmounted by a flag-staff of height h. At a point on the plane, angles of elevation of bottom and top of flag are 45° and 60°. Find h.
Let distance from point to tower = d. tan45° = 20/d → d = 20 m. tan60° = (20+h)/20 → √3 = (20+h)/20 → 20+h = 20√3 → h = 20(√3 − 1) = 20(1.732 − 1) = 14.64 m.
Q4Angle of elevation of airplane from point A is 60°. After 10 seconds flying at same height, elevation becomes 30°. Speed is 720 km/hr. Find the constant height.
Speed = 720 km/hr = 200 m/s. Distance in 10s = 2000 m. Let height = h, horizontal distance from A to first position = x. tan60° = h/x → x = h/√3. tan30° = h/(x + 2000) → x + 2000 = h√3. h/√3 + 2000 = h√3 → 2000 = h√3 − h/√3 = h(3−1)/√3 = 2h/√3. h = 1000√3 m.
Q5From a point 30 m away from a tower, the angle of elevation of the top is 60°. Find the height.
tan60° = h/30. √3 = h/30. h = 30√3 m.
Practice Some Applications of Trigonometry
Reinforce what you just revised with practice questions