A plane is observed approaching your home and you assume its speed to be 550 miles per hour. The angle of elevation of the plane is 16 degrees and one minute later is 57 degrees. Approximate the altitude of the plane.
Draw yourself a figure with two triangles. Both have an apex at your location and a common vertical side going striaght up from there a distance h, where h is the altitude. In one mionute, the plane flys horizontally a distance V t = 550 mile/hr*1/60 hr = 9.17 miles.
h cot 16 - h cot 57 = 9.17 miles
Solve for h.
I get h = 3.23 miles.
Therefore, the altitude of the plane is approximately 3.23 miles.
To approximate the altitude of the plane using the given information, follow these steps:
1. Draw a figure with two triangles. Both triangles should have an apex at your location and a common vertical side going straight up from there a distance h, representing the altitude.
2. In one minute, the plane flies horizontally a distance Vt=550 miles/hour * 1/60 hour = 9.17 miles.
3. Use the trigonometric identities to set up an equation. The difference in the angle of elevation (57 degrees - 16 degrees) corresponds to the change in the vertical distance, which is the altitude h. Use the identity cot(angle) = adjacent/hypotenuse to set up the equation:
h * cot(16 degrees) - h * cot(57 degrees) = 9.17 miles
4. Solve the equation for h. Plug in the values for cot(16 degrees) and cot(57 degrees) from a calculator or trigonometric table.
5. Calculate h using the equation:
h = 9.17 miles / (cot(16 degrees) - cot(57 degrees))
6. Evaluate the expression and round the result to the desired level of precision. From the given information, I obtain h ≈ 3.23 miles as the approximate altitude of the plane.
To approximate the altitude of the plane, you can use trigonometry. Here's how you can solve for it step by step:
1. Draw a figure with two triangles. Both triangles have your location as the apex and a common vertical side going straight up. Let's label this height as 'h', which represents the altitude of the plane.
2. In one minute, the plane travels horizontally a distance of Vt = 550 miles/hr * 1/60 hr = 9.17 miles. This forms the base of one of the triangles.
3. The angle of elevation of the plane changes from 16 degrees to 57 degrees. This forms the angle between the vertical side (height 'h') and the base (9.17 miles).
4. Now, let's focus on one of the triangles. We can use the tangent function to relate the angle and the altitude: tan(angle) = opposite/adjacent. In this case, tan(16 degrees) = h/9.17 miles.
5. Similarly, for the other triangle formed with the angle of 57 degrees, we have tan(57 degrees) = h/(9.17 miles + 9.17 miles). Since the plane travels horizontally for one minute, the base of this triangle will be 2 times the distance traveled.
6. Rearrange both equations to solve for h:
- For the first triangle, you have h = 9.17 miles * tan(16 degrees).
- For the second triangle, you have h = 2 * 9.17 miles * tan(57 degrees).
7. Simplify both expressions for h and subtract them to find the altitude of the plane:
h = (9.17 miles * tan(16 degrees)) - (2 * 9.17 miles * tan(57 degrees))
≈ 3.23 miles
Therefore, the approximate altitude of the plane is 3.23 miles.