Two vehicles p and q leave a station at the same time.p travels 8km/hour on a bearing of 025 degree and q 5km/hour on a bearing of 45 degree west of north.how far apart are they and what is the bearing of p from q after 3hours?

Use the law of cosines. After t hours, the distance d between the vehicles is

d^2 = (8t)^2 + (5t)^2 - 2(8t)(5t)cos70°

because the angle between the bearings if 25+45 = 70°

So, just plug in t=3 and solve for d.

As for the bearing from q to p, just find the slope of the line joining p and q. The angle with that slope is θ, and the bearing is 90-θ degrees east of north.

To find out how far apart the two vehicles are, we can use the concept of components.

First, let's find the distance covered by vehicle p after 3 hours. Since it travels at a speed of 8 km/hour, the distance covered is calculated by multiplying the speed by the time: 8 km/hour * 3 hours = 24 km.

Now, let's find the distance covered by vehicle q after 3 hours. It travels at a speed of 5 km/hour, so the distance covered is calculated as: 5 km/hour * 3 hours = 15 km.

Next, let's calculate the components of the distances traveled by each vehicle.

For vehicle p, the vertical component (north-south) is given by 24 km * sin(25°) = 10.26 km (approx.).
The horizontal component (east-west) is given by 24 km * cos(25°) = 21.84 km (approx.).

For vehicle q, the vertical component (north-south) is given by 15 km * sin(45°) = 10.61 km (approx.).
The horizontal component (east-west) is given by 15 km * cos(45°) = 10.61 km (approx.).

Now, we can find the overall north-south distance by subtracting the vertical components of the two vehicles: 10.61 km - 10.26 km = 0.35 km (approx.).
Similarly, we can find the overall east-west distance by adding the horizontal components of the two vehicles: 10.61 km + 21.84 km = 32.45 km (approx.).

Using the Pythagorean theorem, we can find the overall distance between the two vehicles. The distance is given by sqrt((north-south distance)^2 + (east-west distance)^2):
Distance = sqrt((0.35 km)^2 + (32.45 km)^2) = sqrt(0.1225 km^2 + 1054.9025 km^2) = sqrt(1055.025 km^2) = 32.46 km (approx.).

Therefore, the two vehicles are approximately 32.46 km apart.

Now, let's calculate the bearing of p from q after 3 hours. To do this, we can use trigonometry.

The bearing of p from q is the angle between the line connecting the two vehicles and the north direction.

First, let's find the angle between the line connecting the vehicles and the north direction.

The angle is given by: atan((north-south distance) / (east-west distance)).
Angle = atan(0.35 km / 32.45 km) = atan(0.0108) = 0.62° (approx.).

Next, let's calculate the bearing of p from q.

Since p is at a bearing of 025° and the angle between the line connecting the vehicles and the north direction is 0.62° (counter-clockwise), we can calculate the bearing of p from q as follows:

Bearing of p from q = 025° - 0.62° = 24.38° (approx.).

Therefore, after 3 hours, the two vehicles are approximately 32.46 km apart, and the bearing of p from q is approximately 24.38°.