The problem consists of two parts. Part 1: A girl starts from point A and walks 285m to B on a bearing of $078^{\circ}$. She then walks due south to a point C which is 307m from A. We need to find the bearing of A from C and the distance BC. Part 2: The bearing of a house from point A is $319^{\circ}$. The house is 317m from a point B, which is due east of A. The bearing of the house from B is $288^{\circ}$. We need to find how far the house is from A.

1. Problem Description

The problem consists of two parts.

Part 1: A girl starts from point A and walks 285m to B on a bearing of $078^{\circ}$. She then walks due south to a point C which is 307m from A. We need to find the bearing of A from C and the distance BC.

Part 2: The bearing of a house from point A is $319^{\circ}$. The house is 317m from a point B, which is due east of A. The bearing of the house from B is $288^{\circ}$. We need to find how far the house is from A.

2. Solution Steps

Part 1:

Let's denote the angle between AB and the North direction as $\theta$. Given bearing of B from A is $78^{\circ}$, then $\theta = 78^{\circ}$. Let's create a right triangle AXB where X is on the north line from A, so $\angle BAX = 78^{\circ}$.

Let Y be a point such that C is south of B and AY = 307m. Let us drop a perpendicular from A to BC at point D.

Then, AD = $285 \times \sin(78^{\circ})$

BD = $285 \times \cos(78^{\circ})$

DC = BC - BD

By cosine rule in triangle ABC:

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

$\angle BAC = 180 - 78 = 102^{\circ}$

$285^2 + 307^2 - 2(285)(307) \cos(\angle BAC) = 285^2 + 307^2 - 2(285)(307) \cos(102)$

$285^2 = 81225$

$307^2 = 94249$

$81225+94249 - (174780) \times (-0.20791) = 81225+94249 + 36339.77 = 211813.77$

$BC = \sqrt{211813.77} \approx 460.23 m$

AD = 285sin(78) = 278.28

BD = 285cos(78) = 59.02

CD = $\sqrt{307^2 - (285\sin(78))^2} - 285cos(78)$

$\sin \angle DAC = \frac{285sin(78)}{307}$

Bearing of A from C = $180 + arctan(AD/CD) = \theta$

CD = $\sqrt{AC^2-AD^2} -BD$

CD = $\sqrt{307^2 - (285sin(78))^2}-285cos(78)$

$= \sqrt{307^2-(278.28)^2}-59.02$

$=\sqrt{94249-77437.15} - 59.02$

$=\sqrt{16811.85} - 59.02$

$=129.66-59.02 = 70.64$

$\angle DCA = arctan(278.28/70.64)=75.7$

Bearing of A from C = $180+75.7 = 255.7$

Part 2:

Let A be (0,0). Since B is due east of A and AB is unknown.

B = (x,0). The house H has coordinates (h1,h2).

Bearing of H from A is $319^{\circ}$.

$arctan(\frac{h1}{h2}) = 319-270 = 49^{\circ}$

$h2/h1 = tan 49 \rightarrow h2 = h1tan49$

$h1^2+h2^2= AH^2=AH^2$

$\sqrt{h1^2+h2^2} =$ AH

AH: $\sqrt{h1^2 + h2^2}= \sqrt{h1^2 + h1^2tan^249} = h1\sqrt{1+tan^249}$

$BH=317$

Bearing of house from B is $288^{\circ}$

$arctan(h1-x /h2)=288-270=18$

$h2=tan 18(h1-x)$

So we have

AH: $\sqrt{(x-h_1)^2+h_2^2}$

$h_2/(h_1-x)=tan 18$

AH = $\sqrt{h_1^2+h_2^2}$

$BH^2 = (h_1-x)^2+h_2^2 = 317^2=100489$

arctan $h2/h1 = 360-319$ =41

$319^{\circ}$

$288-270=18$

$x = AB$

$AH=y$

Bearing of H from A is

3

1

9. $y \sin (319) = -y \sin(41) = h1$

$y \cos (319) = y \cos(41) = h2$

$AB = x$

Bearing of H from B is

2

8

8. $-h2/(x+h1) = \tan (18) = - ycos41 / (x-ysin41)$

$y^2 = 317 + x^2$

3. Final Answer

Part 1: The bearing of A from C is approximately $255.7^{\circ}$ and the distance BC is approximately 460.23m.

Part 2: Cannot be solved accurately with information provided. A more detailed diagram and approach would be needed to solve this.