Trigonometry question, Please help!?

Question

Still helping the children! Thankgoodness this is these last 5 questions posted are the last questions that I need help with to help them! The stuff they have to learn today makes my eyes cross! Please put all of the work and any formula you had to do as part of your answer, I need it to understand and then further explain it to him! Thankyou for all of the help!

An aircraft flying at an altitude of 2000 m is approaching an airport. If the angle of depression of the airport is 5°, what is the distance from the plane to the airport measured along the ground? Round your answer to the nearest tenth of a kilometre.

Answer 1

Not sure what you mean by angle of depression, but will assume if is the angle of the line of sight pf the airport relative to plumb normal. If so, tan(5degrees) = x/2000 --> x= 174.977...= 0.174977...km = 0.2 km, rounded to 0.1 km. But it really depends on what angle is the reference here.

If the angle is relative to parallel (same as the angle from the airport to the plane relative to the ground), we get tan(5) = 2000/x --> x = 22.9 km (rounded).

Answer 2

I think more clarification is in order here. Imagine the plane flying in a Cartesian coordinate plane, where its changing position with respect to the airport is given by the coordinates (x, y) at any given time. Then the distance the plane is from the airport horizontally, or along the ground, is represented by the x-axis of the coordinate system, and the vertical distance the plane is from the airport is given by its flight altitude, y, on the y-axis. The angle of depression is the angular measure between any flight passenger's horizontal line of sight and the diagonal line connecting the airport and the passenger's eye. Three distinct points of intersection are formed: one where the line of sight meets a vertical line perpendicular to and extending directly above the airport; another where the diagonal line and the vertical line meet at the airport, and the last where the diagonal line meets the line of sight at the eye. These points of intersection form a right triangle which allow us to use the tangent function of an angle, because we have the angle Θ = 5° and the vertical distance separating the plane and the airport. Knowing these two variables enable us to solve for the third variable used in the tangent function of this problem, the horizontal distance separating the airport and the airplane.

The tangent of any angle is given by its vertical change divided by its horizontal change, so tan Θ = y/x.

We know y = 2000 m and the angle, Θ = 5°. We can solve for x by first manipulating the equation algebraically, and then substituting Θ and y into the equation above (tan Θ = y/x) to find x's actual value. First we solve for x algebraically:

x = y / tan Θ

Now we substitute in known values and solve for the unknown value of x:

y = 2,000 m
tan 5° = 0.08749

x = 2,000 m / 0.087488663
x = 22,860.10474 m

One tenth of a kilometre is 100 metres, so rounding this to the nearest tenth of a kilometre, the answer is closest to 22,900 m, which is equal to 22.9 km.

If you want the answer to the closest tenth of a metre, not kilometre, then the answer rounds to 22,860.1 m.

Answer 3

OK!
Since I can't attach any graphics to this answer,I have to ask you to imagine this figure:

I assume the ground is a straight line(cause the 2000 meter altitude is too small compared to the radius of the earth).

So there is a triangle formed here:
Corner A is the airport
Corner B is the plane
Corner C is a point on the ground just beneath the airplane.
Obviously this triangle is a right triangle cause the angel C is a right angel.(If you need more explanation on this you can email me)
since the angel of depression is 5, then the angle by which the airplane reaches the airport runway is 5.
According to the definitions,
Tangent of angel " a " is Sin(a)/Cos(a)
Sin(a) = the side of the triangle opposite the angel a / hypotenuse
Cos(a) = the side of the triangle near the angel a / hypotenuse.
So Tan a = the side of the triangle opposite the angel a/ the side of the triangle near the angel a

So in here:
Tan 5 = 2km / horizontal distance
Tan 5 = 0.087488663
So the horizontal distance =2 / 0.087488663 and therefor:
horizontal distance = 22.9 km

It is great to have people like you around that value the children's education.Keep going my friend. Good Luck!

Answer 4

Draw a triangle such that :-
PB = 2000 is vertical
BA is horizontal
Angle PAB = 5°
BA = x
tan 5° = 2000 / x
x = 2000 / tan 5°
x = 2286 m is required answer.

Answer 5

tan(5°) = 2,000/x where x is the ground distance (Hint: Draw a picture ☺) So
x = 2,000/tan(5°) = 22,860.1 m

Doug