angle of depression?

Question

hey i was having trouble with these two probs i have the answers but i wasn't sure if its correct, will you please check and if i do get it wrong will you explain the probs to me?
Thanks a bunch love.

1.) distance between towns.
A passenger in an airplane at an altitude of 10 kilometers sees two towns directly to the east of the plane. The angles of depression to the towns are 28 degrees and 55 degrees. how far apart are the towns?
Answer i got was 12.21 km.

2.) angle of depression.
Acellular telephone tower that is 150 feet tall is placed on toop of a mountain that is 1200 feet above sea level. What is the angle of depression from the top of the tower to a cell phone user who is 5 horizontal miles away and 400 feet above sea level?
Answer i got was 2.06 degrees.

Answer 1

1)
10km/d1 = tan(28)
10km/d1 = 0.5317
d1 = 10km/0.5317
d1 = 18.81 km

10km/d2 = tan(55)
10km/d2 = 1.428
d2 = 10km/1.428
d2 = 7.00 km

Difference is (d1 - d2), which is 11.81. You got 12.21 so we don't quite agree.

2)
((150' + 1200') - 400') / 5 miles = tan(d)
950'/26400' = tan(d)
d = tan^-1(0.035985)
d = 2.06 degrees

Answer 2

Question 1
Let plane be at point B
BA is vertical, A at ground level.
AC = x (horizontal)
CD = y (horizontal)
AD = x + y (horizontal)
Angle ABC = 35°
Angle ABD = 62°
tan 35° = x / 10
x = 10.tan 35°
x = 7 km
tan 62° = (x + y) / 10
18.8 = x + y
y = 11.8
Towns are 11.8 km apart

Question 2
Sketch vertical line thro` A,B,C and D where:-
AB = 150 ft
BC = 800 ft
CD = 400 ft
D is at sea level
Draw horizontal DE = 5 miles
Vertical FE = 400 ft
FC is horizontal.
tan FAC = (5 x 5280) / 950
Angle FAC = 87.9°
angle of depression = 2.1°

Answer 3

1) is found by 10 / tan (28) - 10/tan(55) . I got 11.80 km

2) is found by seeing what angle has a tangent of (1200+150-400) over (5 * 5280) I got 2.06 like you did.

Answer 4

Answer (2) is correct: arctan(95/2640) = 2.06deg.

(1)
If A is the town with angle of depression 55deg, and B is the town with angle of depression 28deg, P is the position of the plane and S is the point on the ground vertically below P, then:
From triangle SPA:
SA= 10cot(55deg)
SB = 10cot(28deg)
AB = SB - SA = 10(cot(28) - cot(55))
= 10(1.8807 - 0.7002)
= 11.805km.

Answer 5

1)       ------------
           |\   28
           |  \
           |   \
10km |62\
           |       \
           |        \
           |          \
           -----------
                 x

x is the horizontal distance east from the plane to the first town. Similiar concept for the second town.

Ergo the distance between them is 10km*(tan62-tan35) = 11.81km. I guess you can assume perfectly level terrain.

2)      *-----------
          |\ 90-x
          | \
950ft |x \
          |     \
          |      \
          |        \
          |          \
         ------------
            5mi -> 5*5280ft = 26400ft

tan x = 26400/950
x = tan­­­­­­­¯¹ (26400/950)
x~ 87.94 deg
90-x ~ 2.06 deg

Answer 6

1) tan(90-28) = d1/10

d1 = 18.807km

tan(90-55) = d2/10

d2 = 7.002km

d1 - d2 = 11.8km


2) tan(90 - x) = (26400ft)/ (1200+150-400)

90 - x = 87.94

x = 2.06 degrees

Answer 7

you've the following a good triangle which include the tower, the floor, and the line between Erin and her father. the attitude of melancholy as seen through Erin is an same because the attitude of elevation as seen through her father. She seems down at him on an same attitude as he seems up at her. the attitude on the daddy is the attitude whose tangent is four hundred/2 hundred, an same because the attitude of melancholy at Erin.

Answer 8

You shouldn't have changed your avitar. I was just being sarcastic. Sorry.