Martes, Marso 20, 2012



Objectives:

       At the end of the lesson, the students are expected to:
              a. define the following terms:
                     Angle of elevation
                     Angle of depression
                     Line of sight
b. apply sine, cosine and tangent ratios to find angles of elevation and depression;
c. measure lengths and use measurements to determine angle measures;
              d. solve word problems regarding angle of elevation and depression.








Angles of Elevation / Inclination and
     Angles of Depression / Declination














The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).







If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the angle of depression.







The picture below illustrates an example of an angle of depression and an angle of elevation 






























Images that shown "Angle of Elevation"












































































Images that shown "Angle of Depression"

































Here below are some examples to illustrate the concepts and application as stated above:


Example 1: Given- you are standing on the top of the building, where from the angle of elevation of top of a 120 ft tower is 10 degrees. From a window 6ft below the top of the building, the angle of depression of the base of the tower is 30 degrees. Find out the height of the building and the distance between the tower and the building.

Step 1: Draw the sketch NTS) and insert information given (below):
Step 2: Note that (a + b + 6) =120

Step 3: tan 10 = a/d, and tan 30= b/d

Step 4: From above, a = d* tan 10° and b= d* tan 30 Plug these values of a and b, in expression at step 2 above.

Step 5: It simplifies to: (d * tan 10+ d * tan 30° + 6) = 120
Thus d = 114/( tan 10+tan 30)= 151 ft, as the final answer.(values of tan 10 & tan 30 are obtained from trig tables or use calculator).
Example 2: Given- if a plane that is flying at an altitude of 35,000 feet wants to land at JFK, it must begin its descent so that the angle of depression to the airport is 8. Find out the how many miles from the airport must the plane start descending?
Step 1: The altitude is 35000 ft and angle of depression is 8(Given)

Step 2: tan 8= 35000/d, (the distance is assumed d, from air port to the point where from plane starts descending.)
From the trig tables tan 8= 0.1405, therefore, d= 35000/ 0.1405= 249110 ft or 249110/5280 = 47.2 miles (use measurement conversion- 1 mile = 5280 ft), as the final answer.  

Example 3: Given- Is the angle of elevation and angle of depression numerically are equal?

Step 1: Recall and notice that the angle of elevation and the angle of depression are the interior alternate angles of two horizontal parallel lines.


Step 2: From the theorem learn earlier for parallel lines and transversal, you know that if a transversal intersects two parallel lines; the interior alternate angles are equal.
Therefore, the angle of elevation and angle of depression numerically are equal, is the final answer.


Example 4: Given- Harry (2 m tall) stands on horizontal ground 20 m from a tree. The angle of elevation of the top of the tree from his eyes is 26°. Calculate the height of the tree.

Step 1: Say, the height of the tree be ‘h’. Sketch a diagram to insert the given information.

Step 2: tan 26= (h - 2)/20.
Simplifying it gives, (h - 2) = 20 * tan 26=  20 * 0.4877 = 9.75 m (value obtained from trig table for tan or you may use calculator to find it.)
Thus the height of the tree is 9.75 m (rounded to two decimal places), as the final answer.
 
Example 5: Given- a ranger's tower is located 50m from a tall tree. The angle of elevation to the top of the tree is 10°from the top of the tower, and the angle of depression to the base of the tree is 20°. How tall is the tree?

Step 1: Draw the sketch and insert the given information.

Step 2: tan 20= Tower Ht/50.
So tower height = tan 20* 50 = 0.3640 * 50= 18.2 m
Step 3: Say, height of the tree is h m.
So, (h - 18.2)/50 = tan 10 (0.1763 from tri table).
Thus, h -18.2 = 50 * 0.1763 = 8.82. h = (8.82 + 18.2) = 27 m (rounded)
The height of the tree is 27 m, as the final answer.
 


http://www.winpossible.com/lessons/Geometry_Angles_of_Elevation_and_Depression_-_Geometry.html


The videos below will explain more in detail about Angles of Elevation and Depression, and how to apply the concepts in solving real-world problems. This is explained with the help of several examples and done watching video. This helps you to deal with solving problems and help doing the Trigonometry home work.



















Check Your Understanding






 Work Sheet
  1. A balloon at a height of 31 meters from the ground level is attached to a string inclined at 57o to the horizontal. Find the length of the string to the nearest meter.
a.
37 meters
b.
57 meters
c.
40 meters
d.
34 meters


 2. A ladder with its foot on a horizontal flat surface rests against a wall. It makes an angle of 30° with the horizontal. The foot of the ladder is 41 ft from the base of the wall. Find the height of the point where the ladder touches the wall.
a.
24 ft
b.
23 ft
c.
26 ft
d.
29 ft


 3. Simon stands at 170 m away from the base of a building of height 79.22 m. Find the angle of depression of Simon from the top of the building.
a.
30°
b.
25°
c.
46°
d.
35°

4. The angle of elevation of the top of a dowel from a point on the ground is 36°. After walking 36 ft towards the dowel, the angle of elevation becomes 49°. What is the height of the dowel rounded to the nearest foot?
a.
76 ft
b.
71 ft
c.
74 ft
d.
69 ft


 5. A 37 ft high tree standing vertically upwards is broken by the wind in such a way that its top just touches the ground and makes an angle of 52° with the ground. At what height from the ground did the tree break?
a.
20 ft
b.
16 ft
c.
14 ft
d.
18 ft


 6. A man on the deck of a ship is 15 ft above sea level. He observes that the angle of elevation of the top of a cliff is 70° and the angle of depression of its base at sea level is 50°. Find the height of the cliff and its distance from the ship.

a.
23 ft and 23 ft
b.
238 ft and 41ft
c.
50 ft and 13 ft.
d.
33 ft and 20 ft.


 7. A jet when 3400 m high passes vertically above another jet at an instant when the angles of elevation of the two jets from the same point on the ground are35° and 20° respectively. Find the vertical distance between the two jets to the nearest meter.

a.
1630 m
b.
1638 m
c.
1633 m
d.
1643 m

 8. Two trees of equal height stand on either side of a roadway, which is 50 ft wide. At a point on the roadway between the trees, the elevations of the tops of the trees are 40° and 25°. Find the height of the trees.


a.
22 ft
b.
15 ft
c.
18 ft
d.
20 ft


 9. The angle of elevation of the top of a tree is 30o from a point 28 ft away from the foot of the tree. Find the height of the tree rounded to the nearest feet.


a.
23 ft
b.
10 ft
c.
16 ft
d.
8 ft


 10. A flagstaff stands on the top of a building. At a distance of 48 ft away from the foot of the building, the angle of elevation of the top of a flagstaff is 60° and the angle of elevation of the top of a building is 46°. Find the height of the flagstaff to the nearest feet.

a.
22 ft
b.
17 ft
c.
16 ft
d.
34 ft

 11. The angle of elevation of the top of a hill from the foot of a tower is 65° and the angle of elevation of the top of the tower from the foot of the hill is 50°. If the distance between the foot of the tower and the foot of the hill is 75 ft, then find the height of the tower and the height of the hill rounded to the nearest feet.


a.
161 ft & 89 ft
b.
90 ft & 162 ft
c.
89 ft & 161 ft
d.
162 ft & 90 ft


 12. The angle of elevation of the top of a tower from a point on the ground is 60°. If the height of the tower is 131 m, then find the distance of the point from the foot of the tower.


a.
1313√ m
b.
131 m
c.
3√131 m
d.
1313√ m


 13. From the top of a spire of height 50 ft, the angles of depression of two cars on a straight road at the same level as that of the base of the spire and on the same side of it are 25° and 40°. Calculate the distance between the two cars.


a.
47 ft
b.
7.013 ft.
c.
40.523 ft.
d.
27.786 ft


 14. A boy is standing on the ground and flying a kite with a string of length 210 ft at an angle of elevation of 30°.Another boy is standing on the roof of a 85 ft high building and is flying his kite at an angle of elevation of 55°. Both the boys are facing each other. What shall be the length of the string of the kite of the second boy so that the two kites would touch?
a.
26 ft
b.
24 ft
c.
28 ft
d.
29 ft


 15. The angle of elevation of a cloud observed from a point at a height 170 ft above the level of water in a lake is 54°. The angle of depression of its image in the lake from the same point is 67°. Find the height of the cloud above the lake.



a.
804 ft
b.
814 ft
c.
806 ft
d.
809 ft


 16. From the top of a building of height h meters in a street, the angles of elevation and depression of the top and the foot of another building on the opposite side of the street are θ° and φ° respectively. Find the height of the opposite building.[Given h = 70, φ = 35, and θ = 50.] 

a.
214 meters
b.
224 meters
c.
229 meters
d.
219 meters


 17. From the top of a tower of height h meters the angle of depression of a red car is φ° and the angle of depression of a blue car which is on the opposite side of the tower is θ°. If the distance between the foot of the tower and the red car is y meters, then find the height of the tower and the distanc between the two cars. 
Assume that the points of location of the cars and the foot of the tower are collinear.[Given y = 115, θ = 55° and φ = 40°.]

a.
96 m & 182 m
b.
98 m & 183 m
c.
96 m & 184 m
d.
101 m & 182 m

 18. Two towers are constructed on a plot of land. The angle of depression of the top of the second tower when seen from the top of the first tower is 40°. If the height of the first tower is 135 ft and the height of the second tower is 78 ft, then find the horizontal distance between the two towers.

a.
74 ft
b.
69 ft
c.
79 ft
d.
94 ft


 19. A flagpole of 10 m length is installed on the top of an office building 29 m high. The building and the pole subtend equal angles at a point outside the building which is at a height of 39 m. What is the distance of this point from the top of the pole?

a.
19 m
b.
16 m
c.
14 m
d.
11 m


 20. The angle of elevation of an unfinished tower from a point 120 m away from its base is 25°. How much higher the tower be raised so that its angle of elevation from the same point will be 40°?

a.
50 m
b.
47 m
c.
48 m
d.
45 m

1. A photographer using a camera photographs a rare bird roosting on a high branch of a tree at an angle of elevation of φ°. The distance between the camera lens and the bird is l ft. In order to get a clear picture, the photographer moves closer to the base of the tree. The angle of elevation of the bird is now θ°. Find the new distance between the camera lens and the bird. [Given l = 90, φ = 30° and θ = 55°.]
a.
57 ft
b.
60 ft
c.
62 ft
d.
55 ft


 22. From a shipmast head 135 meters high, the angle of depression of a boat is observed to be 50°. Find the distance of the boat from the ship.
a.
290 m
b.
113 m
c.
161 m
d.
130 m


 23. A telegraph pole is 120 m high. Its shadow is207.972 m in length. Find the angle of elevation of the sun.
a.
33°
b.
20°
c.
40°
d.
30°

 24. The angle of elevation of the top of a cliff from the point Q on the ground is 30°. On moving a distance of20 m towards the foot of the cliff the angle of elevation increases to φ°. If the height of the cliff is17.3 m, then find φ.
a.
45°
b.
60°
c.
120°
d.
30°

 25. Tom and Sam are on the opposite sides of a tower of160 meters height. They measure the angle of elevation of the top of the tower as 40° and 55°respectively. Find the distance between Tom and Sam.


a.
308 m
b.
313 m
c.
303 m
d.
306 m

 26.  A man on the deck of a ship is 13 ft above water level. He observes that the angle of elevation of the top of a cliff is 40° and the angle of depression of the base is 20°. Find the distance of the cliff from the ship and the height of the cliff if the base of the cliff is at sea level.
a.
36 ft and 45 ft.
b.
36 ft and 43 ft.
c.
18 ft and 43 ft.
d.
18 ft and 21 ft


 27. What is the angle of depression in the figure?

a.
Ð


 28.  Is the angle of elevation numerically equal to the angle of depression?

a.
Yes
b.
No


 29. The angles of elevation of the top of a tower from the top and bottom of a 60 m high building are 30o and 60o. What is the height of the tower?




a.
80 m
b.
90 m
c.
55 m
d.
60 m

Chick " Magbasa pa nang higit pa
" to show solutions and references..Thank you. :)



Solutions:


 1.  
A balloon at a height of 31 meters from the ground level is attached to a string inclined at 57o to the horizontal. Find the length of the string to the nearest meter.
a.
37 meters
b.
57 meters
c.
40 meters
d.
34 meters


Solution:


Draw the figure from the given data.

Height of the ballon from the ground level, AC = 31 meters.

The angle made by the string with the horizontal = 57°

In right triangle ABC, sin 57° = 31x
[Formula.]

⇒ 0.84 = 31x
[Substitute the value of sin 57° ]

⇒ x = 310.84 ≈ 37 meters.
[Use calculator.]

The length of the string is 37 meters.

 2.  
A ladder with its foot on a horizontal flat surface rests against a wall. It makes an angle of 30° with the horizontal. The foot of the ladder is 41 ft from the base of the wall. Find the height of the point where the ladder touches the wall.
a.
24 ft
b.
23 ft
c.
26 ft
d.
29 ft


Solution:


Draw the figure for the given data.

The distance from the foot of the ladder to the foot of the wall = 41 ft
[Given.]

Let h be the height of the point A at which the ladder touches the wall.

From right triangle ABC, tan 30° = h41
Þ h = 41 tan 30° » 24 ft.
[Substitute tan 30° and simplify.]

The ladder touches the wall at a height of 24 ft.

 3.  
Simon stands at 170 m away from the base of a building of height 79.22 m. Find the angle of depression of Simon from the top of the building.
a.
30°
b.
25°
c.
46°
d.
35°


Solution:


Draw the figure for the given data.

Height of the building, PQ = 79.22 m.

Distance of Simon from the base of the building, QR = 170 m.

∠SPR = ∠PRQ
[PS and QR are parallel, the angles of elevation and depresion are equal in measure.]

Consider right triangle PQR, tan R = PQ / QR =79.22170
Þ ∠ R » 25°

So, angle of depression of Simon from the top of the building = 25°

 4.  
The angle of elevation of the top of a dowel from a point on the ground is 36°. After walking 36 fttowards the dowel, the angle of elevation becomes49°. What is the height of the dowel rounded to the nearest foot?
a.
76 ft
b.
71 ft
c.
74 ft
d.
69 ft


Solution:


Draw the figure from the given data.

Let P and Q be the points of observation and y be the height of the dowel.

In right triangle QRS, tan 49° = yx

x = ytan 49° » 0.869 y
[Substitute the value of tan 49°.]
[Use calculator.]

In right triangle PRS, tan 36° = y(36 + x)

36 + x = ytan 36°
[Substitute the value of tan 36° and simplify.]

36 + x = 1.375 y

36 + 0.869 y = 1.375 y
[Substitute the value of x = 0.869 y.]

36 = 0.506 y

y » 71

Therefore, the height of the dowel is 71 ft.

 5.  
A 37 ft high tree standing vertically upwards is broken by the wind in such a way that its top just touches the ground and makes an angle of 52° with the ground. At what height from the ground did the tree break?
a.
20 ft
b.
16 ft
c.
14 ft
d.
18 ft


Solution:


Draw the diagram.

A triangle is formed by the broken part of the tree with the ground.

Let x represent the height of the tree from the ground to the broken point.

In right triangle DAC, sin 52° = x37 - x

0.788(37 - x) = x
[Substitute the value of sin 52° and simplify.]

29.156 - 0.788 x = x

29.156 = 1.788 x

x » 16

The tree is broken at a height 16 ft from the ground.

 6.  
A man on the deck of a ship is 15 ft above sea level. He observes that the angle of elevation of the top of a cliff is 70° and the angle of depression of its base at sea level is 50°. Find the height of the cliff and its distance from the ship.
a.
23 ft and 23 ft
b.
238 ft and 41ft
c.
50 ft and 13 ft.
d.
33 ft and 20 ft.


Solution:


Draw the diagram.

Let PQ represent the height of the cliff.

Let RS represent the height of the ship.

Let y represent the distance between the cliff and the ship.

In right triangle PRS, tan 50° = 15y

y = 15tan 50° » 12.58
[Substitute tan 50° = 1.192 and simplify.]

In right triangle QTS, tan 70° = xTS = xy

x = y (tan 70°)

x = 12.58 (tan 70°)
[Substitute the value of y.]

x » 35
[Substitute the value of tan 70° and simplify.]

PQ = PT + QT = 15 + x = 15 + 35 = 50
[From the diagram.]

Therefore, the height of the cliff is 50 ft and its distance from the ship is 13 ft.

 7.  
A jet when 3400 m high passes vertically above another jet at an instant when the angles of elevation of the two jets from the same point on the ground are 35° and 20° respectively. Find the vertical distance between the two jets to the nearest meter.
a.
1630 m
b.
1638 m
c.
1633 m
d.
1643 m


Solution:


Draw the figure from the given data.

Let the height of the first jet = QR = 3400 meters.

Let the height of the second jet = QS meters.

Let the vertical distance between the two jets, RS =l meters.

In right triangle PQR , tan 35° = QR / PQ =3400PQ
ÞPQ = 4855.7 meters.
[Substitute the value of tan 35° and simplify.]

In right triangle PQS , tan 20° = SQ / PQ =SQ4855.7
ÞSQ » 1767 meters.
[Substitute the value of the tan 20° and simplify.]

From the figure RS = RQ - SQ

= 3400 - 1767 = 1633 meters.

So, the vertical distance between the two jets = 1633 meters.

 8.  
Two trees of equal height stand on either side of a roadway, which is 50 ft wide. At a point on the roadway between the trees, the elevations of the tops of the trees are 40° and 25°. Find the height of the trees.
a.
22 ft
b.
15 ft
c.
18 ft
d.
20 ft


Solution:


Draw the diagram.

Let PQ‾ and RS‾ represent the trees.

Let PR‾ represent the distance between the trees.

In right triangle QPT, tan 25° = hx

h = x tan 25°

x » 2.145h
[Substitute the value of tan 25° and simplify.]

In right triangle SRT, tan 40° = h50 - x

h = (50 - x) tan 40°

h = (50 - x)0.839
[Substitute the value of tan 40° and simiplify.]

h = (50 - 2.145h) 0.839

h = 41.950 - 1.799h

2.79h = 41.950

h » 15

h » 15 ft.

Therefore, the height of the trees is 15 ft.

 9.  
The angle of elevation of the top of a tree is 30o from a point 28 ft away from the foot of the tree. Find the height of the tree rounded to the nearest feet.

a.
23 ft
b.
10 ft
c.
16 ft
d.
8 ft


Solution:

The measure of the angle of elevation from point Ais 30.

In right triangle APQ, tan 30° = h28

⇒ h = 28 × 0.577
[Substitute the value of tan 30°.]
[Use calculator.]

⇒ h ≈ 16 ft
[Simplify.]

So, the height of the tree is 16 ft.

 10.
A flagstaff stands on the top of a building. At a distance of 48 ft away from the foot of the building, the angle of elevation of the top of a flagstaff is 60° and the angle of elevation of the top of a building is 46°. Find the height of the flagstaff to the nearest feet.
a.
22 ft
b.
17 ft
c.
16 ft
d.
34 ft


Solution:


Draw a figure for the given data.

Q is the foot of the building. Let x represent the height of the flagstaff and y represent the height of the building.

In right triangle PQS, tan 46° = y48

y = 48 tan 46° = 48 × 1.03 ≈ 49.44
[Substitute the value of tan 46° .]

In right triangle PQR, tan 60° = x + y48

x + y = 48 tan 60°

x + y = 48 × 1.73
[Substitute the value of tan 60° .]

x + 49.44 = 83.04
[Substitute the value of y.]

x ≈ 34

Height of the flagstaff is 34 ft.





 11.  
The angle of elevation of the top of a hill from the foot of a tower is 65° and the angle of elevation of the top of the tower from the foot of the hill is 50°. If the distance between the foot of the tower and the foot of the hill is 75 ft, then find the height of the tower and the height of the hill rounded to the nearest feet.
a.
161 ft & 89 ft
b.
90 ft & 162 ft
c.
89 ft & 161 ft
d.
162 ft & 90 ft


Solution:


Draw the diagram for the given data.

Let x represent the height of the hill and let yrepresent the height of the tower.

Distance between the foot of the tower and foot of the hill = 75 ft
[Given.]

In right triangle PAB, tan 50° = y75

Þ y = 75 tan 50°
Þ y » 89 ft.
[Substitute the value of tan 50° and simplify.]

In right triangle QPA, tan 65° = x75
Þ x = 75 tan 65°
Þ x » 161 ft.

The height of the hill is 161 ft and the height of the tower is 89 ft.
[Rounded to the nearest whole number.]

 12.  
The angle of elevation of the top of a tower from a point on the ground is 60°. If the height of the tower is 131 m, then find the distance of the point from the foot of the tower.
a.
1313√ m
b.
131 m
c.
3√131 m
d.
1313√ m


Solution:


Draw the figure from the given data.

P is the point on the ground.

Let x be the distance of point P from the foot of the tower.

Height of the tower AB = 131 m
[Given.]

In right triangle PAB, tan 60° = 131x.

x = 131tan 60°

x = 1313
[Substitute the value of tan 60°.]

So, the distance of the point P from the foot of the tower = 1313 m.

 13.  
From the top of a spire of height 50 ft, the angles of depression of two cars on a straight road at the same level as that of the base of the spire and on the same side of it are 25° and 40°. Calculate the the distance between the two cars.
a.
47 ft
b.
7.013 ft.
c.
40.523 ft.
d.
27.786 ft


Solution:


Draw the figure for the given data.

Let P and Q be the positions of the cars and xrepresent the distance between them.

Let y represent the distance of the car at P from the foot of the building.

In right triangle BAQ, tan 25° = 50y

⇒ y = 50tan 25°

⇒ y » 107
[Substitute the value of tan 25° and simplify.]

In right triangle BAP, tan 40° = 50 / AP

AP » 59.594
[Substitute the value of tan 40° and simplify.]

AP + PQ = 59.594 + PQ
[Substitute the value of AP.]

AQ = 59.594 + PQ
[From diagram AP + PQ = AQ.]

y = 59.594 + x

107 = 59.594 + x
[Substitute the value of y.]

x » 47

Therefore, the distance between the cars is 47 ft.

 14.  
A boy is standing on the ground and flying a kite with a string of length 210 ft at an angle of elevation of30°. Another boy is standing on the roof of a 85 fthigh building and is flying his kite at an angle of elevation of 55°. Both the boys are facing each other. What shall be the length of the string of the kite of the second boy so that the two kites would touch?
a.
26 ft
b.
24 ft
c.
28 ft
d.
29 ft


Solution:


Draw the diagram.

P and R are the positions of the boys.

QR represents the height of the building.

S is the position where the two kites meet.

In right triangle PUS, sin 30° = SU210

SU = 210 sin 30°

SU » 105
[Substitute the value of sin 30° and simplify.]

ST = SU - TU = SU - RQ
[From the diagram.]

= 105 - 85 = 20

In right triangle STR, sin 55° = ST / SR

SR = STsin 55° = 20sin 55°

SR » 24
[Substitute the value of sin 55° and simplify.]

Therefore, the length of the string of the second boy should be 24 ft so that the two kites would meet.

 15.  
The angle of elevation of a cloud observed from a point at a height 170 ft above the level of water in a lake is 54° . The angle of depression of its image in the lake from the same point is 67°. Find the height of the cloud above the lake.
a.
804 ft
b.
814 ft
c.
806 ft
d.
809 ft


Solution:


Draw the diagram for the given data.

Let A represent the position of the cloud above the lake at a height x ft.

Let B represent the position of the reflection of the cloud in the lake.

In right triangle PSA, tan 54° = AS / PS= x - 170PS

PS = x - 170tan 54° = 0.726(x - 170)
[Substitute the value of tan 54° and simplify.]

In right triangle PSB, tan 67° = BS / PS =170 + xPS

2.356 PS = 170 + x
[Substitute the value of tan 67° and simplify.]

2.356[0.726(x - 170)] = 170 + x

1.710 x - 400.520 = 170 + x

0.710 x = 570.520

x » 804

The height of the cloud above the lake is 804 ft.

 16.  
From the top of a building of height h meters in a street, the angles of elevation and depression of the top and the foot of another building on the opposite side of the street are θ° and φ° respectively. Find the height of the opposite building.[Given h = 70, φ = 35, and θ = 50.] 
a.
214 meters
b.
224 meters
c.
229 meters
d.
219 meters


Solution:


Draw the figure from the given data.

Let AB represent the first building and CErepresent the opposite building.

CD = AB = 70 meters
[The opposite sides of a rectangle.]

In right triangle ABC, tan 35° = AB / BC

⇒ 0.700 = 70BC
[Substitute the value of tan 35° and simplify.]

⇒ BC ≈ 100 meters

AD = BC ⇒ AD = 100 meters
[Opposite sides of a rectangle.]

In right triangle ADE, tan 50° = DE / AD

⇒ 1.191 = DE100
[From step 7.]

⇒ DE ≈ 119.10 meter
[Substitute the value of tan 50° and simplify.]

Therefore, the height of the opposite building CE =CD + DE

⇒ 100 + 119.10

⇒ ≈ 219 meters.

Therefore, the height of the opposite building is 219 meters.

 17.  
From the top of a tower of height h meters the angle of depression of a red car is φ° and the angle of depression of a blue car which is on the opposite side of the tower is θ°. If the distance between the foot of the tower and the red car is y meters, then find the height of the tower and the distanc between the two cars.
Assume that the points of location of the cars and the foot of the tower are collinear.[Given y = 115, θ = 55° and φ = 40°.]
a.
96 m & 182 m
b.
98 m & 183 m
c.
96 m & 184 m
d.
101 m & 182 m


Solution:


Draw the figure from the given data.

Let A represent the top of the tower.

Let BC represents the distance between two cars.

Let h represents the height of the tower and xrepresents the distance between the foot of the tower and the blue car.

In right triangle ABD, tan 40° = AD / BD = h115
Þ h = 115 tan 40°
Þ h » 96 meters.
[Substitute the value of tan 40° and simplify.]

In right triangle ADC, tan 55° = hx
Þ tan 55° = 96y
Þ y » 67 meters.
[Substitute the value of tan 55° and simplify.]

So, the distance between the two cars, BC = 115 +y = 115 + 67 = 182 meters.

 18.  
Two towers are constructed on a plot of land. The angle of depression of the top of the second tower when seen from the top of the first tower is 40°. If the height of the first tower is 135 ft and the height of the second tower is 78 ft, then find the horizontal distance between the two towers.
a.
74 ft
b.
69 ft
c.
79 ft
d.
94 ft


Solution:


Draw the figure for the given data.

Height of the first tower, AB = 135 ft

Height of the second tower, CD = 78 ft

Let the distance between the two towers, BC = ED= x ft.

Let the difference between the heights of the two towers, AE = h ft.

In right triangle AED, tan 40° = AE / ED = hx
⇒ h = 0.83x
[Substitute the value of tan 40° and simplify.]

From the figure, AE = AB - BE
⇒ AE = 135 - 78 = 57 ft.
[From the figure AE = h.]

57 = 0.83x
[From step 6 and step 7.]

⇒ x ≈ 69 ft.

So, the distance between the two towers is 69 ft.

 19.  
A flagpole of 10 m length is installed on the top of an office building 29 m high. The building and the pole subtend equal angles at a point outside the building which is at a height of 39 m. What is the distance of this point from the top of the pole?
a.
19 m
b.
16 m
c.
14 m
d.
11 m


Solution:


Draw the diagram.

Let A represent the position of the point.

Let PQ represent the height of the building.

Let QR represent the height of the pole.

Let x represent the distance of the observer from the top of the pole.

tan θ = 10x
[From right triangle ARQ.]

tan 2θ = 39x
[From right triangle ARP.]

2tan θ1 - tan ² θ = 39x
[Substitute tan 2θ in terms of tan θ.]

2 x tan θ = 39(1 - tan ² θ)

2x(10x) = 39[1 - (10x)²]
[Step 6.]

20 = 39[1 - (10x)²]
[Simplify.]

20 / 39 = 1 - (10x)²

(10x)² = 19 / 39

x² = (10)(10)(39)19
[Solve for x.]

x » 14

Therefore, the distance of the point from the top of the pole is 14 m.

 20.  
The angle of elevation of an unfinished tower from a point 120 m away from its base is 25°. How much higher the tower be raised so that its angle of elevation from the same point will be 40°?
a.
50 m
b.
47 m
c.
48 m
d.
45 m


Solution:


Draw the diagram.

The distance from the base of the unfinished tower to the point B = 120 meters.

Let the height of the tower to be raised to make an angle of 40° at point B be x meters.

Let the height of the unfinished tower, DC = hmeters.

Consider right triangle BCD, tan 25° = DC /BC =h120

Þ h » 55 meters.
[Substitute the value of tan 25° and simplify.]

Consider right triangle ABC, tan 40° = AC /BC =x+h120

Þ x + h = 99.60
Þx + 55 = 99.60
[Substitute the value of tan 40° and simplify.]

Þ x » 45 meters.

So, the height of the tower to be raised to make the angle of elevation 40° at point B = 45 meters.





 21.  
A photographer using a camera photographs a rare bird roosting on a high branch of a tree at an angle of elevation of φ°. The distance between the camera lens and the bird is l ft. In order to get a clear picture, the photographer moves closer to the base of the tree. The angle of elevation of the bird is now θ°. Find the new distance between the camera lens and the bird. [Given l = 90, φ = 30° and θ = 55°.]
a.
57 ft
b.
60 ft
c.
62 ft
d.
55 ft


Solution:


Draw the diagram.

Let B represent the position of the bird.

Let P and Q represent the positions of photographer.

In right triangle PAB, sin 30° = AB / PB= AB / 90

AB = 90 sin 30°

AB = 45
[Substitute the value of sin 30° and simplify.]

In right triangle QAB, sin 55° = AB / BQ = 45 / BQ

BQ = 45sin 55°

BQ » 55
[Substitute the value of sin 55° and simplify.]

Therefore, the new distance between the camera lens and the rare bird is 55 ft.

 22.  
From a shipmast head 135 meters high, the angle of depression of a boat is observed to be 50°. Find the distance of the boat from the ship.
a.
290 m
b.
113 m
c.
161 m
d.
130 m


Solution:


Draw the diagram.

Let BC represent the height of shipmast.

Let A represent the position of the boat.

In right triangle ABC, tan 50° = BC / AB = 135 / AB

AB = 135tan 50° » 113
[Substitute the value of tan 50° and simplify.]

Therefore, the distance of the boat from the foot of the shipmast is 113 m.

 23.  
A telegraph pole is 120 m high. Its shadow is207.972 m in length. Find the angle of elevation of the sun.
a.
33°
b.
20°
c.
40°
d.
30°


Solution:


Draw the diagram.

Let AB represent the height of the pole and AC represent its shadow.

Let θ represent the measure of the angle of elevation.

In right triangle CAB, tan θ = AB / AC = 120207.972= 0.577

θ » 30°

Therefore, the angle of elevation of the sun is 30°.

 24.  
The angle of elevation of the top of a cliff from the point Q on the ground is 30°. On moving a distance of 20 m towards the foot of the cliff the angle of elevation increases to φ°. If the height of the cliff is17.3 m, then find φ.
a.
45°
b.
60°
c.
120°
d.
30°


Solution:


Draw the figure from the given data.

Height of the cliff, PA = 17.3 m.

Distance from point Q to point B = 20 m.

Let BA = x m.

In right triangle PQA, tan 30° = PA / QA =17.320+x

Þ 20 + x = 17.33
Þ x = 10 m.

Consider right triangle PBA, tan θ = PA /BA =17.310

Þ θ = 60°

 25.  
Tom and Sam are on the opposite sides of a tower of160 meters height. They measure the angle of elevation of the top of the tower as 40° and 55°respectively. Find the distance between Tom and Sam.
a.
308 m
b.
313 m
c.
303 m
d.
306 m


Solution:


Height of the pole, AB = 160 m .

Let C & D be the position of Tom and Sam on either side of the tower.

tan 40° = 160BC
[tan R = AB / BC.]

BC » 160tan 40°
[Substitute the value of tan 40° and simplify.]

BC = 190.703 m

tan 55° = 160BD
[tan S = AB / BD.]

BD » 160tan 55° = 112.044 m
[Substitute the value of tan 55° and simplify.]

CD = CB + BD = 190.703 +112.044 » 303 m

So, the distance between Tom and Sam is 303 m.

 26.  
A man on the deck of a ship is 13 ft above water level. He observes that the angle of elevation of the top of a cliff is 40° and the angle of depression of the base is 20°. Find the distance of the cliff from the ship and the height of the cliff if the base of the cliff is at sea level.
a.
36 ft and 45 ft.
b.
36 ft and 43 ft.
c.
18 ft and 43 ft.
d.
18 ft and 21 ft


Solution:


Draw the diagram.

Let PQ represent the height of the cliff.

Let RS represent the height of the ship.

Let y represent the distance between the cliff and the ship.

In right triangle PRS, tan 20° = 13y

y = 13tan 20° » 35.714
[Substitute tan 20° and simplify.]

In right triangle QTS, tan 40° = xTS = xy

x = y (tan 40°)

x = 35.714 (tan 40°)
[Substitute the value of y.]

x » 30
[Substitute the value of tan 40° and simplify.]

PQ = PT + QT = 13 + x = 13 + 30 = 43
[From the diagram.]

Therefore, the distance of the cliff from the ship is 36 ft and the height of the cliff is 43 ft.

 27.  
What is the angle of depression in the figure?
a.
Ð


Solution:

The angle formed by the horizontal line and the line of sight below the horizontal line is called as angle of depression.

In the figure, ∠BAC is the angle of depression.

 28.  
Is the angle of elevation numerically equal to the angle of depression?
a.
Yes
b.
No


Solution:

The angle of elevation and the angle of depression are the interior alternate angles of two parallel lines (horizontal lines).

If a transversal intersects two parallel lines, the interior alternate angles are equal.

So, the angle of elevation and angle of depression are numerically equal.

 29.  
The angles of elevation of the top of a tower from the top and bottom of a 60 m high building are 30oand 60o. What is the height of the tower?

a.
80 m
b.
90 m
c.
55 m
d.
60 m


Solution:

The height of the building DC = 60 m

Angle of elevation of the top of the tower from the top of the building = 30o

Angle of elevation of the top of the tower from the bottom of the building = 60o

Let, x be the distance between building and tower.

tan D = opposite side/adjacent side
[Choose an appropriate trigonometric ratio.]

From ΔADE, tan 30o = AE / DE

0.5773 = hx
x = h0.5773
[Multiply both sides with x / 0.5773.]

From ΔABC, tan 60o = (h+60) / x

1.732 = (h+60)x
[Substitute the values.]

x = (h+60)1.732
[Multiply both sides with x / 1.732.]

From steps (7) and (10), h / 0.5773= (h+60) / 1.732

3h = h + 60
[Multiply both sides with 1.732.]

h = 602 = 30
[Divide both sides by 2 and simplify.]

The height of the tower = h + 60


= 30 + 60 = 90
[Substitute and simplify.]

The height of the tower is 90 m.




References and further readings:









http://www.mathsteacher.com.au/year10/ch15_trigonometry/12_elevation_depression/23elevdep.htm#elevation

http://www.mathwarehouse.com/geometry/angle/elevation_depression/index.php

http://www.winpossible.com/lessons/Geometry_Angles_of_Elevation_and_Depression_-_Geometry.html

http://www.purplemath.com/modules/incldecl.htm

http://neaportal.k12.ar.us/index.php/2010/11/angle-of-elevation-and-depression/

http://commons.bcit.ca/math/competency_testing/testinfo/testsyll11/trigonometry/solvertprobs/angelevdepr/angelevdepr.pdf

http://www.youtube.com/watch?v=-6GthfFMMWE

http://www.wfu.edu/~mccoy/outdoor/aj.pdf     
 
 http://www.teachnet-uk.org.uk/2007%20projects/maths-c2_worksheets/c2%20pure%20maths%20worksheets/Chapters/Chapter%202/Notes-examples/Ch%202.3%20-%20angles%20of%20elevation%20and%20depression.pdf

http://www.eowyn.org/teaching/Trigonometry/Day_08_Notes.pdf

http://campuses.fortbendisd.com/campuses/documents/homework/homework_20100212_1127.pdf  

http://www.youtube.com/watch?v=4vcsrcWRlLs

http://www.youtube.com/watch?v=ouKuKbi-RVY

http://www.youtube.com/watch?v=xhaAf9aK3Kk

http://www.youtube.com/watch?v=oLsZB1VzxQo

http://www.youtube.com/watch?v=-6GthfFMMWE--with







http://worksheets.tutorvista.com/angle-of-elevation-and-depression-worksheet.html?page=4


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