A plane is flying east, ascending at a constant angle. Jack is standing on the ground, watching the plane come to (horizontally) directly at him. When he first sees the plane, at an angle of elevation of 15 degrees, its horizontal distance from him is 11,000. When he sees it again, at an angle of elevation of 35 degrees, its horizontal distance from him is now only 7,ooo away.
-What is the angle thatthe plane is ascending at?
-When the plane is directly over jack, What is it's altitude?
-Hint: use similar triangles
find X
Find Y
To solve this problem, we can use similar triangles. Let's label the unknown values as follows:
- Angle of ascent: α
- Altitude when directly over Jack: h
- Distance from Jack to the plane when first seen: x
- Distance from Jack to the plane when seen again: y
Using the information provided in the problem, we have the following:
1. When the plane is first seen:
- Angle of elevation: 15 degrees
- Horizontal distance: x = 11,000
2. When the plane is seen again:
- Angle of elevation: 35 degrees
- Horizontal distance: y = 7,000
Now, let's solve step-by-step:
Step 1: Find the length of the opposite side (altitude) relative to the angle of elevation of 15 degrees.
Using the definition of tangent, we have:
tan(15 degrees) = h / x
h = x * tan(15 degrees)
Step 2: Find the length of the opposite side (altitude) relative to the angle of elevation of 35 degrees.
Using the definition of tangent, we have:
tan(35 degrees) = h / y
h = y * tan(35 degrees)
Step 3: Set the two expressions for h equal to each other since they represent the same value:
x * tan(15 degrees) = y * tan(35 degrees)
Step 4: Solve for the angle of ascent (α), using the fact that the sum of the angles in a triangle is 180 degrees:
α = 180 degrees - 15 degrees - 35 degrees
α = 130 degrees
So, the angle of ascent of the plane is 130 degrees.
Step 5: Find the altitude when the plane is directly over Jack.
Using the equation h = x * tan(15 degrees) (from Step 1), we substitute the value of x = 11,000:
h = 11,000 * tan(15 degrees)
h ≈ 2,931.69
Therefore, when the plane is directly over Jack, its altitude is approximately 2,931.69 units.
Now let's move on to finding x and y:
Step 6: Find x.
Using the equation h = x * tan(15 degrees) (from Step 1), we substitute the value of h = 2,931.69:
2,931.69 = x * tan(15 degrees)
x ≈ 12,857.98
Therefore, the distance from Jack to the plane when first seen is approximately 12,857.98 units.
Step 7: Find y.
Using the equation h = y * tan(35 degrees) (from Step 2), we substitute the value of h = 2,931.69:
2,931.69 = y * tan(35 degrees)
y ≈ 4,781.02
Therefore, the distance from Jack to the plane when seen again is approximately 4,781.02 units.
To summarize:
- The angle of ascent of the plane is 130 degrees.
- When the plane is directly over Jack, its altitude is approximately 2,931.69 units.
- The distance from Jack to the plane when first seen is approximately 12,857.98 units (x).
- The distance from Jack to the plane when seen again is approximately 4,781.02 units (y).
To solve this problem, we will use the concept of similar triangles. Let's label the information we have:
- Angle of elevation at the first sighting: 15 degrees
- Horizontal distance at the first sighting: 11,000
- Angle of elevation at the second sighting: 35 degrees
- Horizontal distance at the second sighting: 7,000
Now, let's break down the problem into different parts:
1. Finding the angle of ascent:
Using the concept of similar triangles, we can set up the following equation:
tan(15 degrees) = X / 11,000
Since we want to find the value of X (the altitude), we can rearrange the equation:
X = tan(15 degrees) * 11,000
So, by calculating the right side of the equation, we can find the angle of ascent.
2. Finding the altitude when the plane is directly over Jack:
At this point, the horizontal distance between Jack and the plane is zero. We can consider this as the base of the right triangle formed by the plane's altitude and the distance from Jack. Let's call the altitude (the height of the triangle) h.
Since we already have the value for X (the altitude), we can write the following equation:
tan(35 degrees) = h / 7,000
Simplifying the equation, we can find the altitude:
h = tan(35 degrees) * 7,000
By calculating the right side of the equation, we can determine the altitude when the plane is directly over Jack.
3. Finding X and Y:
From the given information, we know that the plane is ascending at a constant angle. Therefore, the horizontal distance traveled by the plane from the first sighting to being directly over Jack is the same as the horizontal distance from the second sighting.
Using the concept of similar triangles, we can set up the following equation:
11,000 / X = 7,000 / Y
By cross-multiplying and rearranging the equation, we find:
X = (11,000 * Y) / 7,000
In this case, we want to find X (the altitude) when the plane is directly over Jack. We already calculated the altitude h based on the second sighting, so we can substitute that value into the equation:
X = (11,000 * h) / 7,000
By calculating the right side of the equation, we can determine X.
Finally, to find Y, we can substitute the values we found for X and h into the equation:
Y = (7,000 * X) / (11,000 * h)
By calculating the right side of the equation, we can determine Y.