Creative depiction of two aircrafts flying into the sky, one at an angle of 6 degrees and the other at an angle of 5 degrees. The first aircraft is traveling at a great speed, leaving a trail behind it. Visualize numbers that symbolize speed but without explicit written text. The second aircraft is flying over an imaginary line signifies the one-mile distance it's intended to travel. The sky palette consists of hues of blues and whites symbolizing daylight. In the far horizon, an 8,000 feet imaginary milestone hovers. The ground appears far beneath, giving depth to the image.

1.A plane takes off at an angle of 6 degrees traveling at the rate of 200 feet/second. If it continues on this flight apth at the same speed, how many minutes will it take to reach a n altitude of 8000 feet?

2. A plane takes off at an angle of 5 degrees. After traveling 1 mile along this flight path, how high (in feet) is the plane above the ground?

1. The rate of elevation increase is 200* sin 6 = 20.9 ft/s. Divide 8000 feet by that number for the number of SECONDS it takes. Then convert that to minutes. I get about 6 1/3 minutes.

2. 5280 feet* sin 5

mmmhh, my calculator says 18(tan63) = 35.32

i asking is this answer the right answer though?

tan 63 = height/18

height = 18tan63
= 35.32

yes

thanx much!

good job ladies. that's what i like to see. :)

lolz thanx

To solve both questions, we will use trigonometry and some basic mathematical calculations.

1. To find out how many minutes it will take for the plane to reach an altitude of 8000 feet, we need to calculate the vertical distance covered by the plane.

First, we convert the angle from degrees to radians since trigonometric functions use radians. We can use the formula: radians = degrees * (π/180)

Angle in radians = 6 * (π/180) ≈ 0.1047 radians

Now, we can use the trigonometric function sine (sin) to find the vertical distance traveled by the plane. The formula to calculate the vertical distance is:

Vertical distance = Speed * Time * sin(angle)

Here, the speed of the plane is given as 200 feet/second and the vertical distance is 8000 feet. We need to find the time.

So, the formula becomes:

8000 = 200 * Time * sin(0.1047)

To isolate time, we divide both sides of the equation by (200 * sin(0.1047)):

Time = 8000 / (200 * sin(0.1047))

Time ≈ 23.9516 seconds

To convert seconds to minutes, we divide by 60:

Time ≈ 23.9516 / 60 ≈ 0.3992 minutes

Therefore, it will take approximately 0.3992 minutes for the plane to reach an altitude of 8000 feet.

2. To calculate how high the plane is above the ground after traveling 1 mile, we need to calculate the vertical distance using trigonometry.

We can use the tangent (tan) function to find the vertical distance. The formula is:

Vertical distance = Distance * tan(angle)

Here, the distance traveled by the plane is given as 1 mile (5280 feet), and the angle is 5 degrees.

So, the formula becomes:

Vertical distance = 5280 * tan(5)

Vertical distance ≈ 465.60 feet

Therefore, the plane is approximately 465.60 feet above the ground after traveling 1 mile.

Another question is a guy wire stretches from the top of an antenna tower to a point on level ground 18 feet from the base of the tower. The angle between the wire and the ground is 63 degrees. How high is the tower?

So said the tower was 3503 fet high because i did tan63 * 18; I feel this answer is wrong..if it is can you please give me the right answer and explain why my mehtod was wrong?