A house has a bearing of $319^\circ$ from point A. Point B is 317 meters due east of A. The bearing of the house from point B is $288^\circ$. We need to find the distance between the house and point A.

1. Problem Description

A house has a bearing of $319^\circ$ from point A. Point B is 317 meters due east of A. The bearing of the house from point B is $288^\circ$. We need to find the distance between the house and point A.

2. Solution Steps

Let H be the location of the house.

Let A and B be the two points.

The bearing of H from A is $319^\circ$. This means the angle measured clockwise from North at A to the line AH is $319^\circ$. The angle between the North and AH measured counter-clockwise is $360^\circ - 319^\circ = 41^\circ$.

The bearing of H from B is $288^\circ$. This means the angle measured clockwise from North at B to the line BH is $288^\circ$. The angle between the North and BH measured counter-clockwise is $360^\circ - 288^\circ = 72^\circ$.

Since B is due east of A, the angle between the North line at A and the line AB is $90^\circ$.

In triangle ABH, let $a$ be the length of BH, $b$ be the length of AH, and $c$ be the length of AB, which is given as $c = 317$ meters.

The angle HAB is the angle between the North at A and AB minus the angle between the North at A and AH. Thus, $\angle HAB = 90^\circ - (360^\circ - 319^\circ) = 90^\circ - 41^\circ = 49^\circ$.

The angle ABH is the angle between the North at B and BA minus the angle between the North at B and BH. The angle between the North at B and BA is $180^\circ$ (since B is east of A). Therefore, $\angle ABH = 180^\circ - (360^\circ - 288^\circ) = 180^\circ - 72^\circ = 108^\circ$.

The sum of angles in a triangle is $180^\circ$, so $\angle AHB = 180^\circ - \angle HAB - \angle ABH = 180^\circ - 49^\circ - 108^\circ = 23^\circ$.

Using the Law of Sines, we have:

$\frac{a}{\sin(\angle HAB)} = \frac{b}{\sin(\angle ABH)} = \frac{c}{\sin(\angle AHB)}$

We are looking for $b$, the distance between A and H.

$b = \frac{c \sin(\angle ABH)}{\sin(\angle AHB)} = \frac{317 \sin(108^\circ)}{\sin(23^\circ)}$

$b = \frac{317 \times 0.9511}{0.3907} \approx \frac{301.5}{0.3907} \approx 771.7$

3. Final Answer

The distance between the house and point A is approximately 771.7 meters.