While flying a pilot spots a water tower that is 4.8km away to the north. At the same time, she also sees a monument that is 5.6km away to the south. The tower and the monument are separated by a distance of 7 km along the flat ground. Find the angles at which the pilot views the water tower.
Well, it seems like the pilot has found themselves in quite a puzzling situation! Let's see if we can help them out.
First, we can draw a diagram to visualize the situation. We have a water tower to the north, a monument to the south, and a straight line connecting them on the ground.
Now, let's use some trigonometry to find the angles at which the pilot views the water tower.
Since the pilot is 4.8km away from the water tower, and the distance between the water tower and the monument is 7km, we can use the Law of Cosines to find the angle at the water tower.
The Law of Cosines states that c^2 = a^2 + b^2 - 2ab*cos(C), where c is the side opposite angle C.
In this case, a = 4.8km, b = 7km, and c is the distance from the pilot to the monument, which is given as 5.6km.
So, applying the Law of Cosines, we have:
(5.6km)^2 = (4.8km)^2 + (7km)^2 - 2(4.8km)(7km)*cos(C)
Simplifying, we get:
31.36km^2 = 23.04km^2 + 49km^2 - 67.2km^2*cos(C)
Rearranging the equation, we have:
67.2km^2*cos(C) = 23.04km^2 + 49km^2 - 31.36km^2
Finally, solving for cos(C), we get:
cos(C) = (23.04km^2 + 49km^2 - 31.36km^2) / (67.2km^2)
Now, to find the angle C at which the pilot views the water tower, we take the inverse cosine (or arccos) of cos(C):
C = arccos((23.04km^2 + 49km^2 - 31.36km^2) / (67.2km^2))
Phew, calculations aside, the angle C is what we're after! Keep in mind that this is just for one angle, and to find the other angle, we subtract C from 180 degrees since the total sum of angles in a triangle is 180 degrees.
But hey, wouldn't it be easier for the pilot to just look out the window?
To find the angles at which the pilot views the water tower, we can use trigonometry.
Let's consider the right triangle formed by the pilot, the water tower, and the point directly below the water tower on the flat ground.
Using the given information, we have:
- The distance from the pilot to the tower (adjacent side) = 4.8 km
- The distance from the tower to the point directly below it (hypotenuse) = 7 km
We can use the cosine function to find the angle θ1 at which the pilot views the water tower:
cos(θ1) = adjacent / hypotenuse
cos(θ1) = 4.8 / 7
θ1 = arccos(4.8 / 7)
Similarly, we can consider the right triangle formed by the pilot, the monument, and the point directly below the monument on the flat ground.
Using the given information, we have:
- The distance from the pilot to the monument (opposite side) = 5.6 km
- The distance from the monument to the point directly below it (hypotenuse) = 7 km
We can use the sine function to find the angle θ2 at which the pilot views the water tower:
sin(θ2) = opposite / hypotenuse
sin(θ2) = 5.6 / 7
θ2 = arcsin(5.6 / 7)
Therefore, the angles at which the pilot views the water tower are θ1 = arccos(4.8 / 7) and θ2 = arcsin(5.6 / 7).
To find the angles at which the pilot views the water tower, we can use trigonometry. Let's first draw a diagram to visualize the situation.
We have a water tower located 4.8km away to the north and a monument located 5.6km away to the south. These two points are separated by a distance of 7km along the flat ground. We want to find the angles at which the water tower is viewed by the pilot.
Let's denote the angle at which the pilot sees the water tower as angle A and the angle at which the pilot sees the monument as angle B. To find angle A, we can use the sine function, and to find angle B, we can use the cosine function.
To find angle A:
sin(A) = opposite/hypotenuse
sin(A) = 4.8km/7km
To solve for A, we can take the inverse sine (also known as arcsine) of both sides:
A = arcsin(4.8km/7km)
Now, let's find angle B:
cos(B) = adjacent/hypotenuse
cos(B) = 5.6km/7km
To solve for B, we can take the inverse cosine (also known as arccosine) of both sides:
B = arccos(5.6km/7km)
Calculating the values of angles A and B using a calculator:
A = arcsin(0.6857) ≈ 43.64 degrees
B = arccos(0.8) ≈ 37.4 degrees
Therefore, the angle at which the pilot views the water tower is approximately 43.64 degrees, and the angle at which the pilot views the monument is approximately 37.4 degrees.
Make your sketch and label point P, M, and W as plane, monument and water tower respectively.
Since we have all 3 sides of the triangle, we can use the cosine law to find one of the angles.
Let's find angle M
4.8^2 = 5.6^2 + 7^2 - 2(5.6)(7)cosM
cosM = (31.36 + 49 - 23.04)/78.4
= ...
find M
then use the sine law to find one of the other angles.
After that one step will get you the third angle.
Because of alternate angles in parallel lines, angle W will be your answer.