Angle of depression?

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.

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

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

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

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.

1)聽聽聽聽聽聽聽------------
聽聽聽聽聽聽聽聽聽聽聽|\ 聽 28
聽聽聽聽聽聽聽聽聽聽聽|聽聽\
聽聽聽聽聽聽聽聽聽聽聽|聽聽聽\
10km聽|62\
聽聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽聽-----------
聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽聽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)聽聽聽聽聽聽*-----------
聽聽聽聽聽聽聽聽聽聽|\ 90-x
聽聽聽聽聽聽聽聽聽聽|聽\
950ft聽|x聽\
聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽聽|聽聽聽聽聽聽聽聽聽聽\
聽聽聽聽聽聽聽聽聽------------
聽聽聽聽聽聽聽聽聽聽聽聽5mi -> 5*5280ft = 26400ft

tan x = 26400/950
x = tan颅颅颅颅颅颅颅炉鹿 (26400/950)
x~ 87.94 deg
90-x ~ 2.06 deg

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.

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掳
Angle ABD = 62掳
tan 35掳 = x / 10
x = 10.tan 35掳
x = 7 km
tan 62掳 = (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掳
angle of depression = 2.1掳