an airplane is flying 75km south and 35 km west of an airport. It is flying at a heigh of 6 km
A)what is the straight-line distance to the airport?
B)What is the angle of elevation of the airplane, from the point of of view of the airport?
oops.
θ = arcsin(6/83) = 4.145°
θ = arcsin(6/83) = 4.145°
To solve these questions, we can use the Pythagorean theorem and trigonometric functions. Let's break down each question step by step:
A) To find the straight-line distance to the airport, we can use the Pythagorean theorem. The horizontal distance (west) and vertical distance (south) form a right triangle with the straight-line distance as the hypotenuse.
1. Draw a diagram to represent the problem. Place the airplane at the origin (0, 0) and draw lines to represent the distances traveled. Label the horizontal line (west) as 35 km and the vertical line (south) as 75 km.
2. Use the Pythagorean theorem: a^2 + b^2 = c^2, where a and b are the lengths of the two smaller sides (35 km and 75 km) and c is the hypotenuse (the straight-line distance we're looking for).
c^2 = 35^2 + 75^2
c^2 = 1225 + 5625
c^2 = 6850
3. Take the square root of both sides of the equation to solve for c:
c ≈ 82.77 km
Therefore, the straight-line distance to the airport is approximately 82.77 km.
B) To find the angle of elevation of the airplane from the point of view of the airport, we can use trigonometric functions.
1. In the same diagram, draw a line from the airplane's position to the airport and extend it just above the airplane to represent the height of 6 km.
2. We now have a right triangle formed by the straight-line distance (hypotenuse) and the height (opposite side). We can use the tangent function to find the angle of elevation.
tan(angle) = opposite/adjacent
tan(angle) = height/distance
tan(angle) = 6 km / 82.77 km
3. Use the inverse tangent function (tan^(-1)) to find the angle:
angle ≈ tan^(-1)(6/82.77)
angle ≈ 4.15 degrees
Therefore, the angle of elevation of the airplane from the point of view of the airport is approximately 4.15 degrees.
The plane is at a distance d where
d^2 = 75^2+35^2+6^2 = 6886
d = 83 km (82.99, but who's checking?)
the angle of elevation θ = arctan(6/83) = 4.169° = 4°10'8"