A, B and C are three towns, the bearing of B and C from A being 295° and 205°, and their distances from A are 280km and 800km respectively. Find the bearing of C from B to the nearest minute.
To find the bearing of C from B, we need to consider the bearings of B and C from A.
First, let's draw a diagram:
```
A
\
\
B-------------C
```
The bearing of B from A is 295°, which means it is 295° clockwise from North.
The bearing of C from A is 205°, which means it is 205° clockwise from North.
Now, let's find the bearing of C from B.
Since C is farther away from A than B, we need to subtract the bearing of C from the bearing of B.
295° - 205° = 90°
Therefore, the bearing of C from B is 90°.
To convert this to minutes, we need to remember that there are 60 minutes in a degree.
So, the bearing of C from B to the nearest minute is 90°.
To find the bearing of C from B, we need to use trigonometry and apply the concept of bearings.
First, let's draw a diagram to represent the situation:
```
B
/
/
A/
\
\
C
```
The bearing of B from A is given as 295°. This means that if you were standing at point A and looking towards point B, the angle between the line AB and the north direction is 295°. Similarly, the bearing of C from A is 205°.
Now, let's find the bearing of C from B. To do this, we need to find the angle between the line BC and the north direction when looking from point B. To find this angle, we can use the Law of Sines.
First, let's find the length of side BC. We can use the Law of Cosines to find this:
BC² = AB² + AC² - 2 * AB * AC * cos(angle BAC)
Since we know the lengths AB = 280 km and AC = 800 km, and the angle BAC = 295° - 205° = 90°, we can substitute these values into the formula:
BC² = 280² + 800² - 2 * 280 * 800 * cos(90°)
BC² = 78400 + 640000 - 448000 * 0
BC² = 78400 + 640000
BC² = 718400
BC ≈ 847.14 km (rounded to two decimal places)
Now that we know the lengths of sides AB, AC, and BC, we can use the Law of Sines to find the angle between line BC and the north direction at point B (let's call it angle BCB'). The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant.
sin(angle BCB') / BC = sin(angle ACB) / AC
We know AC = 800 km, BC = 847.14 km, and angle ACB = 205°, so let's substitute these values into the formula:
sin(angle BCB') / 847.14 = sin(205°)/800
Now, solve for sin(angle BCB'):
sin(angle BCB') = (847.14 * sin(205°)) / 800
sin(angle BCB') ≈ 0.8668
Using an inverse trigonometric function (such as arcsine, sin^(-1)), we can find the angle BCB' in radians and then convert it to degrees:
angle BCB' ≈ arcsin(0.8668) ≈ 59.22°
Therefore, the bearing of C from B is approximately 59.22° to the nearest minute.
Did you make your sketch.
After I filled in the angles I found angle A to be 90°
so you have a right-angled triangle.
Yeahhh, Pythagoras!
I can find the other angles!
tanB = 800/280 = ...
angle B = appr 70.7°
then angle C = 90-70.7 = 19.3°
state the result as a "bearing"