A space shuttle pilot flying toward the Suez Canal finds that the angle of depression on one end of the canal is 38.25deg and the angle of depression to the other end is 52.75deg. If the canal is 100.6 mi long, find the altitude of the space shuttle.
Tan52.75 = h/d. d = hor. dist. from
shuttle to canal.
h = d*Tan52.75 = Altitude of shuttle.
Tan38.25 = h/(d+100.6)
h = (d + 100.6)* Tan38.25
Substitute d * Tan52.75 for h:
d*Tan52.75 = (d + 100.6)*Tan38.25
Solve for d:
1.3151d = 0.7883d + 79.31
1.3151d - 0.7883d = 79.31
0.5268d = 79.31
d = 79.31/0.5268 = 150.6mi = Hor.
dist. from shuttle to canal.
h = 150.6*Tan52.75 = 198mi = Alti-
tude of shuttle.
Well, that space shuttle pilot must be having a "canal"-tastic view from up there! Let's solve this math problem in a fun way.
First, let's imagine the space shuttle pilot floating in the air, looking down at the Suez Canal. From one end of the canal, they have an angle of depression of 38.25 degrees. Now, think of this angle as the pilot trying to look down to the bottom of the canal without getting a “crick” in their neck.
Similarly, from the other end of the canal, the pilot has an angle of depression of 52.75 degrees. At this point, they are probably wondering if someone played a "canal trick" on them!
Since the pilot is flying in a straight line, we can consider the two angles of depression as forming a straight line with the space shuttle in the middle. If we draw a line connecting the two ends of the canal, it will make a nice "canal path."
Now, the length of the canal is given as 100.6 miles, which means the "canal path" is also 100.6 miles long.
To find the altitude of the space shuttle, we need to determine the height of the "canal path." This is where a bit of math comes into play.
Let's call the altitude of the space shuttle "h”. Now, we have a right triangle with the following information:
Angle of depression at one end = 38.25 degrees
Angle of depression at the other end = 52.75 degrees
Length of the "canal path" = 100.6 miles
Guess what? We can use some trigonometry to solve this! Specifically, we can use the tangent function:
Tangent of an angle = Opposite / Adjacent
In this case, the opposite side is the altitude of the space shuttle, "h," and the adjacent side is half of the "canal path," which is 100.6 miles / 2.
Let's plug in the values:
tangent(38.25 degrees) = h / (100.6 miles / 2)
tangent(52.75 degrees) = h / (100.6 miles / 2)
Now, calculating those tangents (with a little help from my "trusty" calculator)...
tangent(38.25 degrees) ≈ 0.7814
tangent(52.75 degrees) ≈ 1.3080
Now, we set up two equations:
0.7814 = h / (100.6 miles / 2)
1.3080 = h / (100.6 miles / 2)
Simplifying the equations:
0.7814 = h / 50.3 miles
1.3080 = h / 50.3 miles
Now, we'll cross-multiply:
0.7814 * 50.3 miles = h
1.3080 * 50.3 miles = h
Solving for "h":
h ≈ 39.29 miles
h ≈ 65.79 miles
Well, well! It appears that the altitude of the space shuttle is approximately 39.29 miles or 65.79 miles, depending on which side of the canal you're looking at!
Please keep in mind that these values are approximate. I hope using a bit of humor made this problem more enjoyable. Have a "shuttle"-rific day!
To solve this problem, we can use trigonometry.
Let's assume that the altitude of the space shuttle is 'h' mi.
We can first calculate the distance between the space shuttle and one end of the canal using the tangent of the angle of depression.
tan(38.25°) = h / x
Solving for x:
x = h / tan(38.25°) Equation 1
Similarly, we can calculate the distance between the space shuttle and the other end of the canal:
tan(52.75°) = h / (100.6 - x)
Solving for x:
x = 100.6 - h / tan(52.75°) Equation 2
Since we know the canal is 100.6 mi long, the sum of the distances from the space shuttle to each end should be equal to 100.6 mi:
x + (100.6 - x) = 100.6
Simplifying:
100.6 - x + x = 100.6
x cancels out, and we are left with:
100.6 = 100.6
This equation is true, meaning our assumptions and calculations are correct.
Now, let's substitute x in Equation 2 with the value from Equation 1:
100.6 - h / tan(52.75°) = h / tan(38.25°)
Multiplying both sides by tan(38.25°) * tan(52.75°):
(100.6 * tan(38.25°) * tan(52.75°)) - h = h * tan(38.25°) * tan(52.75°)
Rearranging the equation:
h * (tan(38.25°) * tan(52.75°) + 1) = 100.6 * tan(38.25°) * tan(52.75°)
Dividing both sides by (tan(38.25°) * tan(52.75°) + 1):
h = (100.6 * tan(38.25°) * tan(52.75°)) / (tan(38.25°) * tan(52.75°) + 1)
Calculating the value:
h = (100.6 * 0.7906 * 1.301) / (0.7906 * 1.301 + 1)
h ≈ 40.77 miles
Therefore, the altitude of the space shuttle is approximately 40.77 miles.
To find the altitude of the space shuttle, we can consider the given information about the angles of depression and the length of the canal.
Let's break down the problem into smaller parts and use trigonometry to find the solution step by step.
Step 1: Draw a diagram
Draw a diagram to visualize the problem. Label the points A and B as the two ends of the Suez Canal. Label the point P as the position of the space shuttle in the sky.
|\
| \
h | \ h
| \
| \
--------C---------A-----
D x
Step 2: Identify the relevant trigonometric ratios
In triangle APD, we have an angle of depression of 38.25 degrees. The opposite side is h (altitude) and the adjacent side is x (horizontal distance).
In triangle BPD, we have an angle of depression of 52.75 degrees. The opposite side is h (altitude) and the adjacent side is the same x (horizontal distance).
The trigonometric ratio we can use to find h is tangent (tan):
tan(angle) = opposite/adjacent
Step 3: Set up equations using the trigonometric ratios for each triangle
For triangle APD:
tan(38.25 degrees) = h/x (Equation 1)
For triangle BPD:
tan(52.75 degrees) = h/x (Equation 2)
Step 4: Solve the equations simultaneously
To solve the equations simultaneously, we can eliminate x by setting the right sides of Equation 1 and Equation 2 equal to each other:
h/x = h/x
tan(38.25 degrees)/x = tan(52.75 degrees)/x
tan(38.25 degrees) = tan(52.75 degrees)
Now, we can solve for h.
Step 5: Calculate the altitude (h) of the space shuttle
Using a scientific calculator, find the inverse tangent (arctan) of both sides to solve for h:
h = x*tan(38.25 degrees)
Now, substitute the given length of the canal (x = 100.6 mi) and calculate the altitude:
h = 100.6 * tan(38.25 degrees)
Using a calculator, enter 38.25 degrees and then find the tangent (tan) of the angle. Multiply the result by 100.6:
h ≈ 100.6 * 0.7931
h ≈ 79.87 (rounded to two decimal places)
Therefore, the altitude of the space shuttle is approximately 79.87 miles.