Forces of 15 kg and 10 kg act on a point with an angle of 60◦ between the forces. Find the magnitude of the

resultant force (to nearest tenth of a kilogram) and the angle (to the nearest degree) the resultant makes with the 15
kg force

Ann is trying to find the height of a building. She finds that the angle of elevation from the ground to the top of the

building is 38◦. She walks 150 feet straight back and finds the angle of elevation to the top of the building is 25◦.
Find the height to the nearest foot of the building

An airplane is headed northeast, bearing 45◦, with an air speed of 500 miles per hour, with a wind blowing with

bearing 315◦, at a speed of 75 miles per hour. Find the ground speed (to nearest tenth of a mph) and the actual
course of the plane (to nearest tenth of a degree).

An airplane is headed northeast, bearing 45◦, with an air speed of 500 miles per hour, with a wind blowing with

bearing 315◦, at a speed of 75 miles per hour. Find the ground speed (to nearest tenth of a mph) and the actual
course of the plane (to nearest tenth of a degree).

To find the magnitude of the resultant force and the angle it makes with the 15 kg force, we can use vector addition. First, we need to resolve the forces into their horizontal and vertical components.

Let's start by resolving the 15 kg force. The horizontal component (F1x) can be calculated using the formula:

F1x = F1 * cos(θ1)

Where F1 is the magnitude of the force and θ1 is the angle it makes with the horizontal axis. Given that the angle between the forces is 60 degrees, the angle between the 15 kg force and the horizontal axis (θ1) is 30 degrees. So we have:

F1x = 15 kg * cos(30°) ≈ 12.99 kg

The vertical component (F1y) can be calculated using the formula:

F1y = F1 * sin(θ1)

So:

F1y = 15 kg * sin(30°) ≈ 7.5 kg

Now let's resolve the 10 kg force. Since the angle between the forces is 60 degrees, the angle between the 10 kg force and the horizontal axis (θ2) is 90 - 60 = 30 degrees. Therefore, we have:

F2x = 10 kg * cos(30°) ≈ 8.66 kg
F2y = 10 kg * sin(30°) ≈ 5 kg

Now we can calculate the horizontal and vertical components of the resultant force:

FRx = F1x + F2x
≈ 12.99 kg + 8.66 kg
≈ 21.65 kg

FRy = F1y + F2y
≈ 7.5 kg + 5 kg
≈ 12.5 kg

Using these components, we can find the magnitude of the resultant force (FR) using the Pythagorean theorem:

FR = sqrt(FRx^2 + FRy^2)
≈ sqrt((21.65 kg)^2 + (12.5 kg)^2)
≈ sqrt(470.12 kg^2 + 156.25 kg^2)
≈ sqrt(626.37 kg^2)
≈ 25 kg (rounded to the nearest tenth)

To find the angle (θ) the resultant force makes with the 15 kg force, we can use trigonometry. The angle can be calculated using the formula:

θ = arctan(FRy / FRx)

So:

θ ≈ arctan(12.5 kg / 21.65 kg)
≈ 32.1° (rounded to the nearest degree)

Therefore, the magnitude of the resultant force is approximately 25 kg, and the angle it makes with the 15 kg force is approximately 32 degrees.