A girl starts from point A and walks 285m to B on a bearing of 078.She then walks due south to a point C which is 307m from A. What is the bearing of A from C,and what is {BC}?

please i need the answer urgent ooo it is an assignment pls help me fast

Why can't this solutionr be written properly

Please I need the answer ooo is an assignment please help me

Why can't this solutions be written properly

the girls walks on a heading, not a bearing.

B's bearing from A is 78°

If A is at (0,0), then B is at (279,59)

If we mark point D due east of A, on the line BC, then D is at (279,0)

Now, if angle DAC is θ, then

cosθ = 279/307
θ = 25°

So, A's bearing from C is 270+25 = 295°

BC = 59+279sin25° = 177m

To find the bearing of A from C, we first need to find the bearing of C from A and then add 180 degrees to it. Let's break down the problem step by step:

1. First, let's find the bearing of C from A. We know that the girl walked due south from B to C. So, the bearing of C from B would be 180 degrees (south).

2. To find the bearing of C from A, we need to add the bearing of C from B to the bearing of B from A. The bearing of B from A is 078 degrees. Adding 180 degrees to it (bearing of C from B), we get:

078 + 180 = 258 degrees

Therefore, the bearing of C from A is 258 degrees.

Moving on to the second part of the question:

3. To find the length of BC, we can use the Pythagorean theorem. Given that AC is 307m and AB is 285m, we can calculate BC as follows:

BC^2 = AC^2 - AB^2
BC^2 = 307^2 - 285^2
BC^2 = 94249 - 81225
BC^2 = 13024
BC ≈ √13024
BC ≈ 114.07m

Therefore, BC is approximately 114.07m.

To summarize:
- The bearing of A from C is 258 degrees.
- The length of BC is approximately 114.07m.