A plane is flying south at an altitude of 1.5km when the navigator sees a town directly ahead at an angle of?

Question

depression of 21 degree. At the same time she sees another town due east at an angle of depression of 18 degree. How far apart are the towns.

ans given without solution behind the book is 6.05km, please provide diagram if possible, thanks in advance.

Answer 1

Let AB be height of plane = 1.5 km
A is at ground level
AC is distance of town (ground level)
/_ BCA = 21°
/_ ABC = 69°
tan 69° = AC / 1.5
AC = 1.5 tan 69°
AC = 3.91

D is at ground level
AD is distance of town (ground level)
/_ ABD = 72°
/_ BDA = 18°
tan 72° = AD / 1.5
AD = 1.5 tan 72 °
AD = 4.62

/_ DAC = 90°
Triangle DAC is in horizontal plane (ie ground level)

DC ² = 3.91 ² + 4.62 ²
DC = 6.05

Towns are 6.05 km apart

Answer 2

There 3 right triangles involved here:
two of the them are standing up
one is lying down

D = distance between the two towns
x = length of the base of the south rt triangle
y = length of the base of the east rt triangle

D^2 = x^2 + y^2
D = sqrt[x^2 + y^2]

Solving for x:
The angle of depression is 21 deg; hence
the angle between the vertical angle and the line of sight from the plane to the south town is 90 deg - 21 deg = 69 deg
tan 69 deg = x/1.5km
x = 1.5km(tan 69 deg)
x = 3.91 km

Solving for y:
The angle of depression is 18 deg.; hence,
the angle between the vertical angle and the line of sight from the plane to the east town is 90 deg - 18 deg = 72 deg

tan 72 deg = y/1.5 km
y = 1.5km(tan 72deg)
y = 4.62 km

D = sqrt[x^2 + y^2]
D = sqrt[3.91^2 + 4.62^2]
D = 6.05 km ANS

Hope I help.

teddy boy