town Q is on a bearing 210°from town P. R is on a bearing 150° from town P and R is east of Q. The distance between R and P is 10km. Find the distance between R and Q.

sketch it. you know in the triangle angle P, and angle Q just from the bearings. I get

P=210-150
Q=60 (QR is due East)

so you have two (three if you consider the entire triangle), one opposite side. Law of sines is useful at this point

Vector Q makes an angle of 60o with the -X-axis.

Vector R makes an angle of 60o with the +X-axis.
Each vector forms a 60-30 right triangle.

Using vector R, Cos60 = X/r = X/10, X = 10*Cos60 = 5 km.
sin60 = Y/r = Y/10, Y = 8.66 km. = Ver. component of R and Q.

Using Vector Q, Tan60 = Y/X = 8.66/X, X = 5 km.
QR = 5 + 5 = 10 km = Distance between R and Q.

To find the distance between R and Q, we need to use trigonometry and the concept of bearing.

First, let's draw a diagram to visualize the problem.

```
P
̷
/ \
10km) \
150°\
R \
̷
\
Q
```
In the diagram, P represents town P, and R represents town R. The line segment connecting P and R is 10 km long. Q is situated to the west of P. Town R is east of Q.

We can see that there is a triangle formed by P, Q, and R. We need to find the length of the side connecting R and Q.

Since we are given bearings, we can use the angle between the line PR (150°) and the line QR (210°) to find the angle QRP.
The sum of the angles in a triangle is 180°, so the angle QRP can be found as follows:

Angle QRP = 180° - Angle PR - Angle QR

Angle QRP = 180° - 150° - 210°

Angle QRP = -180°

As we can see, we obtain a negative angle, which means it needs to be adjusted.

Adding 360° to a negative angle will give us the same angle in the positive form.

Angle QRP = -180° + 360°

Angle QRP = 180°

Now that we have the angle QRP (180°), we can use the Law of Cosines to find the length of side QR.

The Law of Cosines states that c^2 = a^2 + b^2 - 2ab * cos(angle_c)

In our case:
a = 10 km (side PR)
b = unknown (side QR)
angle_c = 180°

Therefore:
b^2 = a^2 + c^2 - 2ac * cos(angle_c)

Plugging in the values, we have:
b^2 = 10^2 + 10^2 - 2 * 10 * 10 * cos(180°)

b^2 = 100 + 100 + 200
b^2 = 400

Taking the square root of both sides, we get:
b = √400
b = 20 km

Hence, the distance between R and Q is 20 km.