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Angles of elevation and depression

Two crucial vocabulary items power this chapter.

Definitions

The angle of elevation of a point PP from an observer at OO is the angle that the line of sight OPOP makes with the horizontal through OO, measured upward. It is used when the object is above the observer.

The angle of depression of a point PP from an observer at OO is the angle that the line of sight OPOP makes with the horizontal through OO, measured downward. It is used when the object is below the observer.

Key facts

  • The angle of elevation of PP from OO equals the angle of depression of OO from PP. They are alternate interior angles between parallel horizontals.
  • All angles are measured from the horizontal, never from the vertical.
  • The horizontal is the level eye-line of the observer.
  • The object's height is measured from the base level (typically the ground).

Setting up a problem

A reliable five-step recipe:

  1. Read the problem twice. Identify the observer and the object.
  2. Draw a diagram. Put the observer at the corner of a right triangle, with the horizontal leg on the ground and the vertical leg as the height.
  3. Mark all given lengths and angles. Use letters for the unknowns.
  4. Pick a trig ratio. Usually tan\tan, because the legs are horizontal and vertical and we know one of them.
  5. Solve. Use the standard-angle table if the angle is 30,45,6030^\circ, 45^\circ, 60^\circ. Verify the answer makes physical sense.

Worked examples

Example 1. The angle of elevation of the top of a tower from a point on the ground, 3030 m away, is 3030^\circ. Find the height of the tower.

Let height =h= h. Then tan30=h/30h=301/3=10317.3\tan 30^\circ = h/30 \Rightarrow h = 30 \cdot 1/\sqrt{3} = 10 \sqrt{3} \approx 17.3 m.

Example 2. From the top of a building 2020 m high, the angle of depression of a car on the road is 3030^\circ. Find the distance of the car from the building.

Let the distance =d= d. The angle of depression is 3030^\circ, so the angle between the line of sight and the horizontal at the top is 3030^\circ.

tan30=20/dd=20/tan30=20334.6\tan 30^\circ = 20/d \Rightarrow d = 20/\tan 30^\circ = 20 \sqrt{3} \approx 34.6 m.

Example 3. A ladder leans against a wall at 6060^\circ to the ground. If the foot of the ladder is 44 m from the wall, how long is the ladder?

The ladder is the hypotenuse, the horizontal is 44 m. cos60=4/LL=4/(1/2)=8\cos 60^\circ = 4/L \Rightarrow L = 4/(1/2) = 8 m.

Example 4. From a point on the ground, the angle of elevation of the top of a flagstaff is 6060^\circ. The flagstaff is 2020 m tall. Find the distance from the observer to the flagstaff.

tan60=20/dd=20/3=(203)/3\tan 60^\circ = 20/d \Rightarrow d = 20/\sqrt{3} = (20\sqrt{3})/3 m.

Example 5. An observer 1.51.5 m tall is 3030 m away from a tower. The angle of elevation of the top of the tower from the eye of the observer is 3030^\circ. Find the height of the tower.

Eye-level is at 1.51.5 m. Effective elevation: height from eye-level. tan30=h/30h=30/3=103\tan 30^\circ = h'/30 \Rightarrow h' = 30/\sqrt{3} = 10\sqrt{3} m.

Total tower height =h+1.5=103+1.518.8= h' + 1.5 = 10\sqrt{3} + 1.5 \approx 18.8 m.

Try it yourself

  1. The angle of elevation of a tower from 100100 m is 4545^\circ. Height?
  2. The angle of depression of a boat from a 5050-m cliff is 3030^\circ. Find the boat's distance from the foot of the cliff.
  3. A kite at height 6060 m is attached to a string of length 120120 m. What is the angle of inclination of the string?
  4. A person on a 2020-m hill sees a tree at angle of depression 4545^\circ. Distance of tree?
  5. A pole casts a shadow 3\sqrt{3} times its own length. Find the sun's angle of elevation.
  6. From the top of a 3030-m building, the angle of depression of a point on the ground is 4545^\circ. Find the distance of the point from the building.
  7. The angle of elevation of the top of a tower from two points aa and bb from the foot of the tower in the same straight line is 3030^\circ and 6060^\circ. (Set up only.)
  8. If tanθ=5/12\tan\theta = 5/12 for the elevation of a 5050-m tower, find the distance from the observer.
  9. A bridge 2525 m above a river makes an angle of depression of 4545^\circ with a boat. Find the horizontal distance from the bridge to the boat.
  10. An observer at sea level sees a lighthouse at an angle of elevation of 6060^\circ. The lighthouse is 3030 m tall. Distance?

Pitfalls / Insight

  • The angle is measured from the horizontal, not the vertical.
  • Eye-level vs. ground level , include the observer's height if asked.
  • Diagram first. Algebra without a diagram is the leading cause of mistakes.

Insight. Heights and distances are just a careful choice of right triangle. Once it's drawn, the rest is one trig ratio away.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Angles of elevation and depression
6 questions · pick the best answer
Q1

Angle of elevation is measured from:

Q2

Angle of depression is used when:

Q3

From PP, the depression of OO is 40°40°. The elevation of PP from OO is:

Q4

Both angles are measured from:

Q5

A tower 1515 m, distance 1515 m. Angle of elevation is:

Q6

Best ratio when horizontal and vertical legs are involved: