A girl starts from point A and walks 285m to point B on a bearing of 078 degrees. She then walks due south to a point C which is 307m from A. We want to find the bearing of A from C and angle BAC. A house has a bearing of 319 degrees from point A. The house is 317m due east of A. We need to find the bearing of the house and the distance from B to the house from A.

1. Problem Description

A girl starts from point A and walks 285m to point B on a bearing of 078 degrees. She then walks due south to a point C which is 307m from A. We want to find the bearing of A from C and angle BAC.

A house has a bearing of 319 degrees from point A. The house is 317m due east of A. We need to find the bearing of the house and the distance from B to the house from A.

2. Solution Steps

Problem 1

Let's analyze the triangle ABC. We know AB = 285m, AC = 307m and the angle between North and AB is 78 degrees. Since the girl walks due south from B to C, we can construct a triangle ABC.

We will use the cosine rule to find angle BAC.

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

Let's call the angle BAC as angle A. The angle ABC is (180-78) + 90 = 102 +90 degrees

Also, the horizontal distance from A to B is $AB \cdot sin(78)$ and vertical distance is $AB \cdot cos(78)$. Since C is directly South of B, $BC = AB \cdot cos(78) + \sqrt{AC^2 - (AB\cdot sin(78))^2}$

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

$BC = 285 \cdot 0.2079 + \sqrt{94249 - 77134.77}$

$BC = 59.25 + \sqrt{17114.23}$

$BC = 59.25 + 130.82$

$BC = 190.07$

Now apply the cosine rule:

$307^2 = 285^2 + 190.07^2 - 2(285)(190.07)cos(angle ABC)$

$94249 = 81225 + 36126.6 - 108340.3 cos(angle ABC)$

$-23102.6 = -108340.3 cos(angle ABC)$

$cos(angle ABC) = 0.2132$

$angle ABC = arccos(0.2132) = 77.7 degrees$

Bearing of A from C: Using sine rule $\frac{BC}{sin(angle BAC)} = \frac{AC}{sin(angle ABC)}$, we can find angle BAC. Also we can solve without the cosine rule using coordinates.

Let A be at (0,0).

Then B is at $(285\cdot sin(78), 285\cdot cos(78)) = (278.28, 59.24)$.

Since C is due south from B, the x coordinate is the same.

So C is at $(278.28, 59.24-BC)$.

Also, the distance from A to C is

3

0

7. $AC^2 = 278.28^2 + (59.24-BC)^2 = 307^2$

$77439.4 + 3509.3 -118.48BC + BC^2 = 94249$

$BC^2 -118.48BC -13200.25= 0$

$BC = \frac{118.48 + \sqrt{118.48^2 + 4(13200.25)}}{2}$

$BC = \frac{118.48 + \sqrt{14037.5 + 52801}}{2} = \frac{118.48 + 258.5}{2} = 188.49$

So C is at (278.28,-129.25)

$angle BAC = arctan(\frac{278.28}{-129.25}) = arctan(-2.153) = -65.1$

Bearing of A from C:

Bearing is $arctan(\frac{0-278.28}{0--129.25}) = arctan(\frac{-278.28}{129.25}) = arctan(-2.153) = -65.1$, since angle is in 2nd quadran , $180-65.1 = 114.9$, so bearing = $180+65.1 =245.1$. The bearing is $180 + arctan(278.28/129.25) = 180 + 65.1 = 245.1$.

Angle BAC:

$\frac{188.49}{sin(A)} = \frac{307}{sin(ABC)} = \frac{285}{sin C}$

Problem 2

The house is at an angle of 319 degrees from A. The house is east of A.

Then from A, house is $(d\cdot sin(90),d\cdot cos(90))$ but distance is 317m. The house is at (317,0). Bearing of the house is 319 degrees. So

House is at coordinates ($x_h,y_h$).

$arctan(\frac{x_h-0}{y_h-0}) = 319-270 = 49$

$\frac{x_h}{y_h} = tan(49) = 1.15$

$x_h= 1.15 y_h$. Also $\sqrt{x_h^2 + y_h^2} = distance$ from A to H.

3. Final Answer

Problem 1

Bearing of A from C: Approximately 245.1 degrees

Angle BAC: Approximately 65.1 degrees.

Problem 2

Final Answer: The final answer is incomplete due to a lack of complete and clear information in the prompt. More details regarding the location of the house relative to point B are needed.