: 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|
No, the bearing of B from C is 0°, since the walker moved due south from B to C.
In triangle ABC, if point D is due east from A and lies on BC, then we have
BD = 285 sin12° = 59.25
AD = 285 cos12° = 278.77
CD = √(307^2 - 278.77^2) = 128.58
So BC = CD+DE = 187.84
You are correct, my apologies for the mistake. The bearing of B from C is indeed 0° since the walker moved due south from B to C.
In triangle ABC, we can introduce point D which is due east from A and lies on BC. Let's calculate the lengths of BD, AD, CD, and BC:
BD = AB sin(12°) = 285 sin(12°) ≈ 59.25 m
AD = AB cos(12°) = 285 cos(12°) ≈ 278.77 m
CD = √(AC^2 - AD^2) = √(307^2 - 278.77^2) ≈ 128.58 m
Now, to find BC, we need to add CD and DE:
BC = CD + DE = 128.58 m + 59.25 m ≈ 187.84 m.
Therefore, |BC| is approximately 187.84 m.
To find the bearing of A from C, we need to find the angle made between the line segment AC and the north direction. Let's break this down into two steps.
Step 1: Find the bearing of B from A.
The bearing of B from A is 078°. This means we have to rotate 78° clockwise from the north direction to align with AB.
Step 2: Find the bearing of C from A.
Since C is due south of A, the bearing from A to C is 180°.
Now, let's find the bearing of A from C.
Bearing of A from C = Bearing of C from A + 180°
= 180° + 180°
= 360°.
So, the bearing of A from C is 360°.
Next, let's find the length of BC.
Given that AB = 285m and AC = 307m.
We can use the Pythagorean theorem to find BC.
Using Pythagorean theorem:
BC^2 = AC^2 - AB^2
BC^2 = 307^2 - 285^2
BC^2 = 94249 - 81225
BC^2 = 13024
BC = √13024
BC ≈ 114.05 m
Therefore, |BC| is approximately 114.05 m.