Practical use of Trigonometry

Trigonometric knowledge can be used to estimate heights of objects or distances betweens points, for example, height of a mountain or breadth of a river.

Angles of elevation and depression

If a horizontal line is drawn from the eye of the observer (O), and an object P is above this line OX, then POX is called angle of elevation. If an object Q is below the horizontal line OX, then QOX is called angle of depression. Note that from P, the point O is situated at an angle of depression = X'PO = POX =
                    

Bearings of a point

The bearing of a line in horizontal plane is the angle made by it from North-South line. The angle (measured from north in clockwise direction) is generally given in three figures (e.g. 045° instead of 45°). So this kind of measurement is called three figure bearing.
                            

          

However, more frequently, quadrant bearings are used. Here, firstly we see whether the line makes acute angle on North or South side. Then we measure acute angle from North or South towards East or West as the case may be, and use that measure (in degrees) and letter E or W accordingly.
                               

Some useful figures

While solving problems on heights and distances, the following figures should be remembered.
     2 1· 414,3 1·732,5 2·236

sin 0° = 0, sin 15° = (3 -1)/[22 , sin 36° = (5 - 1)/[22]
sin 18° = (5 -1)/4, sin 30° = 1/2, sin 60° = (3)/2 ,sin 75° = (3 -1)/22
sin 90° = 1

Illustrative Examples

Example

A tree casts shadow 3 times its height. Find the angle of elevation of the sun.

Solution

Let BC = h be the height of the tree, so that length of shadow, AB = 3 h. If is the angle of the elevation of the sun, we see from the diagram that
 tan = BC/AB = h/[h3] = 1/3 = tan 30°
            = 30°               

Example

From the top of a tower 60 m high, the angles of depression of the top and bottom of a building are observed to be 30° and 60° respectively. Find the height of the building. Also find the distance between the tower and the building.

Solution

Let AB be the building h meters high, and PQ be the tower 60 m high situated at distance BQ = x meters away.
Then AC = BQ = x, CQ = AB = h,
PC = 60 -h
From BPQ, cot 60° = BQ/PQ = x/60
=> x = 60 cot 60°
From APC, cot 30° = AC/PC = x/(60 -h)
=>    60 -h = x/ cot 30° = 60 cot 60° tan 30° = 60 (1/3)(1/3) = 20
=>   h = 60 -20 = 40       

Thus the building is 40 meters high and is situated 34·64 meters away from the tower.

Example

Two boats leave a place at the same time. One travels 56 km in the direction N 50° E, while the other travels 48 km in the direction S 80° E. What is the distance between the boats?
                                                    

Solution

Let points A and B represent the position of two boats. We have to find AB, and we know that OA = 56 km, OB = 48 km,
AOB = 180° -50° -80° = 50°
Using cosine formula in OAB, we get
cos 50° =[ (56)²+(48)² - (AB)²]/[ 2(56)(48)]
AB² = (56)² +(48)² -2 (56)(48) cos 50°
= 3136 +2304 -5376 x 0·6424
= 5440 -3455·69 = 1984·31
=> AB = 198.31 = 44·54 meters

Example

A radio transmitter antenna of height h stands at the top of a tall building. At a point on the ground, the angle of elevation of the bottom of the antenna is and that of the top of the antenna is . Prove that the height of the building is h tan /(tan -tan).
                                                   

Solution

Let AB be the building of height x, CA be the antenna of height h, and y be the distance of the observer from the foot of the building.
In OBC, tan = (x +h)/y
In OAB, tan = x/y
Eliminating y between these two equations,
(x +h)/x = tan / tan  =>    x tan +h tan = x tan
=>    h tan = x tan -x tan
=>   x  = h tan /(tan -tan)

Exercise

  1. A kite with a string 150 ft. makes an angle of 45° with the ground. Assuming that the string is straight, how high is the kite?
  2. A tree 10 meters high casts a 17·3 meter shadow. Find the angle of elevation of the sun.
  3. From the top of a hill, the angles of depression of two consecutive kilometre stones due east are 30 and 45 degrees. How high is the hill?
  4. A 12 meter ladder is inclined to the vertical at angle 15°. How far is it from the base?
  5. An observer in a lighthouse is 66 feet above the surface of the water. The observer sees a ship and finds the angle of depression to be 0·7°. Estimate the distance of the ship from the base of the lighthouse. Round the answer to the nearest 5 feet.
  6. From the point on a ground level, you measure the angle of elevation to the top of a mountain to be 45°. Then you walk 200 m further away from the mountain and find that the angle of elevation is now 30°. Find the height of the mountain. Round the answer to the nearest meter.
  7. A surveyor stands 30 yards from the base of a building. On top of the building is a vertical radio antenna. Let denote the angle of elevation when the surveyor sights to the top of the building. Let denote the angle of elevation when the surveyor sights to the top of the antenna. Express the length of the antenna in terms of the angles and .
  8. A helicopter hovers 800 feet directly above a small island. From the helicopter, the pilot takes a sighting to a point directly ashore on the mainland, at the waters edge. If the angle of depression is 30°, how far off the coast is the island?
  9. A ladder 18 feet long leans against a building. The ladder forms an angle of 60° with the ground.
    (i) How high up the side of the building does the ladder reach?
    (ii) Find the horizontal distance from the foot of the ladder to the base of the building.
  10. Two satellite tracking stations, located at points A and B in a desert, are 200 miles apart. At a prearranged time, both stations measure the angle of elevation of a satellite as it crosses the vertical plane containing A and B. If the angles of elevation from A and from B are and respectively, express the altitude h of the satellite in terms of and .
  11. From the base of a 30 m high building, the angle of elevation of a tower is 60°, and from the top of the building, it is 30°.Find the height of the tower.
  12. A man on the top a building 30 meters high observes a man coming directly towards it at a uniform speed. If it takes 15 minutes for the angle of depression to change from 30° to 45°, how much time will it take after this for the man to reach the base of the building? Round your answer to nearest minute.
  13. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°.If after 10 seconds, the elevation is observed to be 30°, find the uniform speed per hour of the plane.
  14. A man standing south of a lamppost observes his shadow on the horizontal plane to be 24 ft long. On walking eastwards 300 ft, he finds his shadow as 30 ft. If his height is 6 ft, find the height of the lamp above the plane.
  15. Two poles of equal height are standing opposite to each other on either side of a road, which is 30 ft wide. From a point between them on the road, the angles of elevation of the tops are 30° and 60°. Find the height of each pole, rounded to nearest feet.
  16. A vertical tower is surmounted by a vertical flagstaff of height 10 ft. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are and respectively. Prove that the height of the tower is 10 tan /(tan -tan) feet.
  17. An aeroplane when 6000 m high passes vertically above another plane at an instant when their angles of elevation at the same observing point are 60° and 45° respectively. How many meters higher is the one than the other?
  18. At the foot of a mountain, the elevation of its peak is 45°. After ascending 100 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain (to the nearest meter).
  19. A vertical pole is struck by a speeding car and breaks into two, the top striking the ground at an angle of 30° and at a distance of 10 feet from the foot of the pole. Find the total height of the pole.
  20. The breadth of a street between two houses is 9 meters and the angle of depression of the top of one, as observed from the top of the other, which is 12 meters high, is 30°. Find the height of the other house.
  21. A town B is 13 km south and 18 km west of town A. Find the bearing and distance of B from A.

Answers

1. 752 ft                           2. Approx 30°                   3. 2·37 km
4. 3·10 m                             5. 5400 ft                          6. 273·22 m
7. 30 (tan -tan)                  8. 8003m
9. (i) 93 ft  (ii) 9 ft              10. 200 /(cot -cot) miles
11. 45 m                               12. 41 minutes                   13. 240 3 km/hr
14. 106 ft                              15. 13 feet                         17. 2536 m
18. 137 m                             19. 17·32 ft                        20. 6·8 m
21. 22·2 km, S 54°10' W