A girl starts at point A and walks 285m to point B on a bearing of $078^\circ$. She then walks due south to point C, which is 307m from A.

1. Problem Description

A girl starts at point A and walks 285m to point B on a bearing of $078^\circ$. She then walks due south to point C, which is 307m from A.

2. Solution Steps

Let's denote the distance from B to C as $x$. Let the angle between AB and AC be $\theta$. We can form a triangle ABC.

First, let's find the angle BAC. Since the bearing of B from A is $078^\circ$, the angle between the north direction from A and the line AB is $78^\circ$. The girl walks due south from B to C. Let D be the point such that AD points north and B is on the line AD. Since we are looking for the bearing of A from C.

We can find the angle between AB and AC using the cosine rule. In triangle ABC, we have:

$BC = x$

$AB = 285$

$AC = 307$

$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)$

$AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(\angle ACB)$

$BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)$

We can use the cosine rule to find the angle $\angle BAC$.

$x^2 = 285^2 + 307^2 - 2 \cdot 285 \cdot 307 \cdot \cos(\angle BAC)$

Also, consider the east-west displacement from A to B. The east component is $285 \cdot \sin(78^\circ)$.

The north component is $285 \cdot \cos(78^\circ)$.

Since C is south of B, the east component of AC is the same as the east component of AB, which is $285 \sin(78^\circ)$.

Let y be the south displacement from A to C.

$y = BC + 285 \cos(78^\circ)$

We also know that $AC = 307$, so

$307^2 = (285 \sin(78^\circ))^2 + (BC + 285 \cos(78^\circ))^2$

$307^2 = (285 \sin(78^\circ))^2 + x^2 + 2 \cdot x \cdot 285 \cos(78^\circ) + (285 \cos(78^\circ))^2$

$307^2 = 285^2 (\sin^2(78^\circ) + \cos^2(78^\circ)) + x^2 + 2 \cdot x \cdot 285 \cos(78^\circ)$

$307^2 = 285^2 + x^2 + 2 \cdot x \cdot 285 \cos(78^\circ)$

$307^2 - 285^2 = x^2 + 2 \cdot x \cdot 285 \cos(78^\circ)$

$94249 - 81225 = x^2 + 2 \cdot x \cdot 285 \cos(78^\circ)$

$13024 = x^2 + 117.72x$

$x^2 + 117.72x - 13024 = 0$

$x = \frac{-117.72 \pm \sqrt{117.72^2 - 4(1)(-13024)}}{2}$

$x = \frac{-117.72 \pm \sqrt{13858.0 + 52096}}{2}$

$x = \frac{-117.72 \pm \sqrt{65954}}{2}$

$x = \frac{-117.72 \pm 256.81}{2}$

$x = \frac{139.09}{2} = 69.545$

$x = BC \approx 69.55$

$69.55^2 = 285^2 + 307^2 - 2 \cdot 285 \cdot 307 \cdot \cos(\angle BAC)$

$4837.2025 = 81225 + 94249 - 174990 \cos(\angle BAC)$

$4837.2025 = 175474 - 174990 \cos(\angle BAC)$

$174990 \cos(\angle BAC) = 170636.7975$

$\cos(\angle BAC) = 0.97512$

$\angle BAC = \cos^{-1}(0.97512) = 12.87^\circ$

The bearing of C from A is $78^\circ - 12.87^\circ = 65.13^\circ$

3. Final Answer

The bearing of C from A is approximately $065^{\circ}$.