Angles of Elevation and Depression

LESSON 6 Angles of Elevation and Depres- sion

LESSON 6

Now we get a chance to apply all of our newly acquired skills to real-life applications, otherwise known as word problems. Let’s look at some elevation and depression problems. I first encountered these in a Boy Scout handbook many years ago. There was a picture of a tree, a boy, and several lines.

Example 1

tree

How tall is the tree? 5'

11' 30'

Separating the picture into two triangles helps to clarify our ratios.

5 θ 11 X

θ 41 5 x We could write this as a proportion (two ratios), = , 11 41 and solve for X.

ANGLES OF ELEVATION AND DEPRESSION - LESSON 6 59 We can also use our trig abilities.

5 From the boy triangle: tan. θ= = 4545 = 24.44º “ ” 11 θ

x From the large triangle: tan. 2444º = 41

Solve for X. ()41 ().4545 = x 18 .63 = x The tree is 18.63'. When working these problems, the value of the trig ratio may be rounded and recorded, and further calculations made on the rounded value. You may also keep the value of the ratio in your calculator and continue without rounding the inter- mediate step. This may yield slightly different final answers. These differences are not significant for the purposes of this course. It is pretty obvious that an angle of elevation measures up and an angle of depression measures down. One of the keys to being a good problem solver is to draw a picture using all the data given. It turns a one-dimensional group of words into a two-dimensional picture.

Figure 1

depression

elevation We assume that the line where the angle begins is perfectly flat or horizontal.

Example 2 A campsite is 9.41 miles from a point directly below the mountain top. If the angle of elevation is 12º from the camp to the top of the mountain, how high is the mountain? top

mountain campsite 12º 9.41 mi

60 LESSON 6 - ANGLES OF ELEVATION AND DEPRESSION PRECALCULUS You can see a right triangle with the side adjacent to the 12º angle measuring 9.41 miles. To find the height of the mountain, or the side opposite the 12º angle, the tangent is the best choice.

height tanº 12 = 94. 1 mi ()94.t11()an 2º = height (()94. 1 ().2126 = height 2 miles = height

Example 2 At a point 42.3 feet from the base of a building, the angle of elevation of the top is 75º. How tall is the building?

height tanº 75 = 42.'3 ()42.t37()an 5º = height building ()42.3 ()3.7321 = height 157.'87 = height of the building

75º 42.3' Practice Problems 1

1. How far from the door must a ramp begin in order to rise three feet with an 8º angle of elevation?

2. An A-frame cabin is 26.23 feet high at the center, and the angle the roof makes with the base is 53º15'. How wide is the base?

PRECALCULUS ANGLES OF ELEVATION AND DEPRESSION - LESSON 6 61 Solutions 1

1. 2. 3 8º 26.23 X

3 tanº 8 = x 53º15' x tanº 83= X X 3 x = tanº 8 3 53º"15 = 53.º25 x = 26.23 .1405 tan. 53 25º = x = 21.35 ft x 26.23 x = tan. 53 25º 23.26 x = 1.3392 26.23 x = 1.3392 x = 19.59 23x 91.8 ft =

62 LESSON 6 - ANGLES OF ELEVATION AND DEPRESSION PRECALCULUS 6A Answer the questions.

1. isaac’s camp is 5,280 feet from a point directly beneath Mt. Monadnock. What is the hiking distance along the ridge if the angle of elevation is 25º 16'?

2. How many feet higher is the top of the mountain than his campsite?

Express as a fraction.

3. csc q = 6. csc a =

4. sec q = 7. sec a = 2 31 4

θ 5. cot q = 8. cot a = 6 3

Express as a decimal.

9. sin q = 12. sin a =

10. cos q = 13. cos a =

11. tan q = 14. tan a =

PRECALCULUS Lesson 6A 53 LESSON 6A

15. Use your answers from #9–11 to find the measure of q.

16. Use your answers from #12–14 to find the measure of a.

Solve for the lengths of the sides and the measures of the angles.

17. α A 12

35.4º B

18. 29º 59 D

α C

19. F 42.66º

E 100

20. G α

47 H 41º32'10''

54 PRECALCULUS 6B Answer the questions.

1. The side of a lake has a uniform angle of elevation of 15º 30'. How far up the side of the lake does the water rise if, during the flood season, the height of the lake increases by 7.3 feet?

2. A building casts a shadow of 110 feet. If the angle of elevation from that point to the top of the building is 29º 3', find the height of the building.

Express as a fraction.

3. csc q = 6. csc a =

4. sec q = 7. sec a =

5. cot q = 8. cot a = α 11 4.6

θ Express as a decimal. 10

9. sin q = 12. sin a =

10. cos q = 13. cos a =

11. tan q = 14. tan a =

PRECALCULUS Lesson 6B 55 LESSON 6B

15. Use your answers from #9–11 to find the measure of q.

16. Use your answers from #12–14 to find the measure of a.

Solve for the lengths of the sides and the measures of the angles.

17. J α 12 18º K

18. M α L 59 29º

19. 67º N 10.25 θ P

20. 6 θ

2 13 Q α

56 PRECALCULUS 6C Answer the questions.

1. From a point 120 feet from the base of a church, the angles of elevation of the top of the building and the top of a cross on the building are 38º and 43º respectively. Find the height to the top of the cross. (The ground is flat.)

2. Find the height of the building as well as the height of the cross by itself.

Express as a fraction.

3. csc q = 6. csc a =

4. sec q = 7. sec a = α 15 5. cot q = 8. cot a = 7.1

θ 13.2 Express as a decimal.

9. sin q = 12. sin a =

10. cos q = 13. cos a =

11. tan q = 14. tan a =

PRECALCULUS Lesson 6C 57 LESSON 6C

Results for #15 and 16 may vary slightly from the solutions, depending on when steps were rounded.

15. Use your answers from #9–11 to find the measure of q.

16. Use your answers from #12–14 to find the measure of a.

Solve for the lengths of the sides and the measures of the angles.

17. S 40º

R 25

18. α T 88

36.2º U

19. α V 150

51.9º W

20. α

95 X

θ 7

58 PRECALCULUS 6D Answer the questions.

1. A campsite is 12.88 miles from a point directly below Mt. Adams. If the angle of elevation is 15.5º from the camp to the top of the mountain, how high is the mountain?

2. At a point 60.7 feet from the base of a building, the angle of elevation from that point to the top is 64.75º. How tall is the building?

Express as a fraction.

3. csc q = 6. csc a =

4. sec q = 7. sec a =

5. cot q = 8. cot a = α 25 X

Express as a decimal. θ 18.33 9. sin q = 12. sin a =

10. cos q = 13. cos a =

11. tan q = 14. tan a =

PRECALCULUS Lesson 6D 59 LESSON 6D

15. Use your answers from #9–11 to find the measure of q.

16. Use your answers from #12–14 to find the measure of a.

Solve for the lengths of the sides and the measures of the angles.

17. 2.24 α

2 Y

18. 10.5 α

A Z 49.2º

19. 29.07º

B 56

α C

20. α

D 10

θ 14

60 PRECALCULUS 6H Here are some more applications of trig functions. In some of these you may need to find a missing side, and in others a missing angle.

Use the skills you have learned so far to answer the questions. Always begin by making a drawing and labeling the known information.

1. A girl who is 1.6 meters tall stands on level ground. The elevation of the sun is 60° above the horizon. What is the length of her shadow?

2. if the girl in #1 casts a shadow that is one meter long, what is the elevation of the sun?

3. A stairway forms an angle with the floor from which it rises. This angle may be called the angle of inclination. What is the angle of inclination of a stairway if the steps have a tread of 20 centimeters and a rise of 16 centimeters?

Some problems will require more of your algebra skills. There are some examples of these on the next page. The first one is done for you.

PRECALCULUS Honors 6H 61 Honors 6H

4. An observation balloon is attached to the ground at point A. On a level with A and in the same straight line, the points B and C were chosen so that BC equals 100 meters. From the points B and C, the angle of elevation of the balloon is 40º and 30º respectively. Find the height of the balloon.

First, make a drawing. There’s not enough information x to find x using either the angle at B or the angle at C.

40º 30º However, we can make two equations using x and y. A y B 100mC

x x Equation 1 tan 40º = Equation 2 tan 30º = y y +100

Replace tan 40º with its ratio and solve for x in Equation 1. .8391 = x or x = .8391y y x Replace tan 30º with its ratio in in Equation 2. .5774 = y +100 .8391y Substitute value of x from Equation 1 in Equation 2. .5774 = y +100

Solve for y. .5774(y + 100) = .8391y .5774 y + 57.74 = .8391y 57.74 = .2617y y = 220.6 (rounded)

Solve for x, which is the height of the balloon. x = .8391y x = .8391 (220.6) = 185.1 m

5. Tom wished to find the width of a river. He observed a tree directly across the river on the opposite bank. The angle of elevation to the top of the tree was 32º. Then Tom moved directly back from the bank 50 meters and found that the angle of elevation to the top of the tree was 21º. What is the width of the river?

6. in the side of a hill that slopes upward at an angle of 32º, a tunnel is bored sloping downward at an angle of 12º15' from the horizontal. How far below the surface of the hill is a point 38 meters down the tunnel?

62 PRECALCULUS t e s t 6 Use for #1–4: Devan stands 926 meters from Use for #5–8: From a point 80 meters from the a point directly below the peak of a mountain. base of a building to the top of the building, the The angle of elevation between Devan and the angle of elevation is 51°. From the same point to top of the mountain is 42°. the top of a flag staff on the building, the angle of elevation is 54°.

1. Which equation can be used to find 5. What equation can be used to find the height of the mountain (x)? the combined height (y) of building and flagpole? A. sin 42º = x/926 B. tan 42º = 926/x A. y = 80 tan 51º C. cos 48º = 926/x B. y = 80 sin 54º D. tan 42º = x/926 C. y = 80 tan 54º D. y = tanº 51 80

2. What is the height of the mountain? 6. What is the height of the building alone? A. 833.8 m B. 1,028.4 m A. 98.8 m C. 619.6 m B. 110.1 m D. 1,383.9 m C. 64.8 m D. 58.1 m 3. A tower 50 meters high is built on top of the mountain. What is the angle of elevation from Devan’s position to the top of the tower? 7. What is the height of the flagpole (Round decimal degrees to tenths.) alone?

A. 40º 14' 44'' A. 15.1 m B. 43º 42' B. 45.3 m C. 57º 15' C. 4.2 m D. 46º 20' 08'' D. 11.3 m

4. If a bird flew from Devan’s position 8. How long must a cable be in order to the top of the mountain, how to stretch from the observation many meters would it travel? point to the top of the building?

A. 408.4 m A. 102.9 m B. 1,246.1 m B. 127.1 m C. 1,383.9 m C. 136.1 m D. 1,280 m D. 50.3 m

PRECALCULUS Test 6 15 test 6

Use for #9–10: A car traveled a distance of 100 13. 46º 21' 02'' = feet up a ramp to a bridge. The angle of elevation of the ramp was 10°. A. 46.21º

9. How high was the bridge above B. 46.12º road level? C. 46.35º A. 17.4 ft B. 98.5 ft D. 46.4º C. 10 ft D. 100 ft

10. What is the actual distance from 14. sin α is equal to: cos α the beginning of the ramp to the base of the bridge? A. tan α

A. 575 ft B. cot α B. 98.5 ft C. 89.4 ft C. sec α D. 17.4 ft D. csc α

11. 3 is the ratio for: 15. 1 is equal to: 3 cos α

A. cos 45º A. csc α B. cos 30º B. sec α C. tan 60º C. sin α D. tan 30º D. cos α

12. Arcsin .8192 =

A. 1.22 B. 35º C. 55º D. .9999

16 PRECALCULUS Lesson6A 5,280 1. cosº25 16' = 5,280 1. cosº25 16' = D D Dcosº25 16',= 5 280 Dcosº25 16',= 5 280 5,2280 D = 5,2280 D = cosº25 16' cosº25 16' Df≈ 5,.838 77 t Df≈ 5,.838 77 t M 2. tanº25 16' = M 2. tanº25 16' = 5,280 5,280 M = ()5,,t280 ()an25º'16 M = ()5,,t280 ()an25º'16 Mf≈ 2,.492 09 t Mf≈ 2,.492 09 t 231 31 3. csc θ ==231 31 3. csc θ ==4 2 4 2 231 31 31 3 93 4. sec θ ==231 31 ==31 3 93 4. sec θ ==63 33==333 9 63 33 333 9 63 33 5. cot θ ==63 33 5. cot θ ==4 2 4 2 93 6. csc α = 93 6. csc α = 9 9 31 7. sec α = 31 7. sec α = 2 2 23 8. cot α = 23 8. cot α = 9 9 4 9. sin.θ = 4 ≈ 3592 9. sin.θ = 231 ≈ 3592 231 63 10. coosθ = 63≈ .9333 10. coosθ = 231 ≈ .9333 231 4 11. tan.θ = 4 ≈ 3849 11. tan.θ = 63≈ 3849 63 63 12. sinα = 63≈ .9333 12. sinα = 231 ≈ .9333 231 4 13. cos.α = 4 ≈ 3592 13. cos.α = 231 ≈ 3592 231 Lesson 6A - LESSON 6A 63 14. tan.α = 63≈ 2 5981 14. tan.α = 4 ≈ 2 5981 4 15. arccsin ..3592 ≈ 21 05º 15. arccsin ..3592 ≈ 21 05º 16. arcsin ..9333 ≈ 68 96º LessonLesson6 6AA 16. arcsin ..9333 ≈ 68 96º 5,280 B 1. cosº25 16' = 17. tan5446.º= B D 17. tan5446.º= 12 12 Dcosº25 16',= 5 280 B = ()12 ()tan.54 61º.≈ 689 B = ()12 ()tan.54 61º.≈ 689 5,2280 D 12 = sin.35 4º = 12 cosº25 16' sin.35 4º = A Df≈ 5,.838 77 t A A sin335.º41= 2 A sin335.º41= 2 M 12 2. tanº25 16' = A≈= 12 20.72 A≈= sin.35 4º 20.72 Lesson6A 5,280 sin.35 4º Lesson6A α =−90 35.º45= 46.º M = ()5,,t280 ()an25º'16 α =−90 35.º45= 46.º D5,280 α =− = 1. cosº25 Mf16≈' 2=,.492 09 t M D D 18. siin61º = D 18. siin61º = 59 Dcosº2523161 ',= 531280 59 3. csc θ == D = ()59 ()sinº61 ≈ 51.6 4 2 5,2280 D = ()59 ()sinº61 ≈ 51.6 25º16' D = cosº25 16' c 4. sec 231 co31sº25 3116'3 93 cosº61 = c θ ==5,280== ft cosº61 = 59 63Df≈335,.8383377 3t 9 59 c = ()59 ()cossº61 ≈ 28.6 63 33 c = ()59 ()cossº61 ≈ 28.6 5.2. cotatnºθ25==16' M 2. tanº25 16' = α =−90ºº29 = 61º 4 52,280 α =−90ºº29 = 61º 93 F 6. csc α = M = ()5,,t280 ()an25º'16 19. tan.47 34º = F 9 19. tan.47 34º = 100 Mf≈ 2,.492 09 t 100 F = (()100 ()tan.47 34º 7. sec 31 F = (()100 ()tan.47 34º α = 231 31 = 3. csc θ ==2 F ≈ 108.52 4 2 F ≈ 108.52 234 2 8. cot α = 100 4. sec 239 1 31 31 3 93 sin.42 66º = 100 4. sec θ ==== sin.42 66º = e 63 33 333 9 e 9. sin.4 ≈ 3592 θ = esin.42 66º = 1100 23631 33 esin.42 66º = 1100 5. cot θ == 100 4 2 e = 100 ≈ 147.57 634 2 e = sin.42 66º ≈ 147.57 10. coosθ = ≈ .9333 sin.42 66º 93 6. csc α = 231 α =−90º.42 66º.= 47 34º 9 α =−90º.42 66º.= 47 34º 49 11. tan.θ = ≈ 3849 G 31 20. ttan.41 54º = G 7. sec α = 63 20. ttan.41 54º = 47 2 47 63 G≈= ()47 ()tan.41 54º.41 64 12. sinα = 23 ≈ .9333 G≈= ()47 ()tan.41 54º.41 64 8. cot α = 23 α = 231 447 9 cos.41 54º 447 4 cos.41 54º = 447 13. cos.α = 4 ≈ 3592 cos.41 54º = H 9. sin.θ = ≈ 3592 H 231 Hcos.41 54º = 47 231 Hcos.41 54º = 47 63 47 14. tan.α = 63≈ 2 5981 H = 47 ≈ 62.79 10. coosθ = 4 ≈ .9333 H = cos.41 54º ≈ 62.79 231 cos.41 54º 15. arccsin ..3592 ≈ 21 05º α =−90ºº41 32'' 10"º= 48 27'"50 4 α =−90ºº41 32'' 10"º= 48 27'"50 11. tan.θ = 4 ≈ 3849 α =− = 11.16. arcsintan.θ =..9333≈≈384968 96º θ ≈ 41.º54 63 θ ≈ 41.º54 B 17. tan5446.º63= 12. sinα = 12≈ .9333 231 B = ()12 ()tan.54 61º.≈ 689 13. cos.4 ≈ 3592 13. cos.α = ≈123592 sin.35234º1= A 63 14. taAn.siαn=335.º41=≈ 225981 4 12 15. arccsin ..3592A≈=≈ 21 05º 20.72 sin.35 4º 16. arcsin ..9333 ≈ 68 96º α =−90 35.º45= 46.º B PR17.eCalctaulusn5446.º=D 18. siin61º = 12 solutions 281 59 B = ()12 ()tan.54 61º.≈ 689 D = ()59 ()sinº61 ≈ 51.6 12 sin.35 4ºc= cosº61 = A 59 A sin335.º41= 2 c = ()59 ()cossº61 ≈ 28.6 12 A≈= 20.72 α =−90ºº29 =si61n.35º 4º 90 35.º45F 46.º 19. αta=−n.9047 3435º.º=45= 46.º 100 D 18. siin61º = F59= (()100 ()tan.47 34º D = ()F59≈ 108()sinº.6152 ≈ 51.6 c 100 cosisºn.6142 =66º = 59 e esin.42c =66()59º = 1100()cossº61 ≈ 28.6 100 α =−90ºº29e = 61º ≈ 147.57 sin.42 66º F 19. αta=−n.9047º.3442º =66º.= 47 34º 100 G 20. ttan.41 54Fº = (()100 ()tan.47 34º 47 F ≈ 108.52 G≈= ()47 ()tan.41 54º.41 64 100 sin.42 66º = 447 cos.41 54º = e H esin.42 66º = 1100 Hcos.41 54º = 47 100 e = 47 ≈ 147.57 H =sin.42 66º ≈ 62.79 cos.41 54º α =−90º.42 66º.= 47 34º α =−90ºº41 32'' 10"º= 48 27'"50 G 20. θtta≈n.4141.º5454º = 47 G≈= ()47 ()tan.41 54º.41 64 447 cos.41 54º = H Hcos.41 54º = 47 47 H = ≈ 62.79 cos.41 54º α =−90ºº41 32'' 10"º= 48 27'"50 θ ≈ 41.º54