A girl starts from a 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/
All angles are measured CW from +y-axis.
.d = 285*sin78 = 278.8 m. = hor. distance of AB from Y-axis.
h = 285*Cos78 = 59.3 m = ht. of AB.
sinX = 278.8/307.
X = 65o E. of S. = 115o CW. = Bearing of AC.
BC = 59.3 + 307*sin(90-65) = 189 m.
To find the bearing of A from C, we need to determine the angle formed between the direction from C to A and the north direction.
Given that the girl walks from A to B on a bearing of 078°, we know that the angle formed between the north direction and the direction from A to B is 78°.
Now, let's determine the direction from C to A. Since the girl walks due south from B to C, the direction from C to A would be the opposite direction, which is due north.
Therefore, the bearing of A from C is 180°.
To find the length of BC, we can use the Pythagorean theorem since we have a right-angled triangle formed by A, B, and C.
Let's consider triangle ABC:
- AB = 285m (given)
- AC = 307m (given)
Using the Pythagorean theorem:
BC^2 = AB^2 + AC^2
BC^2 = (285m)^2 + (307m)^2
BC^2 = 81225m^2 + 94249m^2
BC^2 = 175474m^2
Taking the square root of both sides:
BC = √(175474m^2)
BC ≈ 418.7m (rounded to one decimal place)
Therefore, the length of BC is approximately 418.7m.
To find the bearing of A from C, we need to determine the angle formed by the line segment AC with the north direction.
First, let's find the distance BC.
We have the information that the girl walked 285m from A to B on a bearing of 078°. This means that the angle between the line segment AB and the north direction is 78°.
Using trigonometry, we can calculate the horizontal and vertical components of AB.
The horizontal component (BC) can be calculated using the formula:
BC = AB × cos(θ),
where θ is the angle between AB and the north direction.
BC = 285m × cos(78°).
BC = 285m × 0.2079 (rounded to four decimal places).
BC ≈ 59.2075m (rounded to four decimal places).
Therefore, the distance BC is approximately 59.2075m.
Next, let's determine the bearing of A from C.
Since the girl walks due south from B to C, the line segment BC is vertical.
As we already determined that BC is approximately 59.2075m, the line segment AC can be split into two line segments: AB (285m) and BC (59.2075m).
Using trigonometry, we can find the angle formed by AC with the north direction:
tan(θ) = BC / AB,
where θ is the angle formed by AC with the north direction.
θ = arctan(BC / AB).
θ = arctan(59.2075m / 285m).
θ ≈ 11.747° (rounded to three decimal places).
Therefore, the bearing of A from C is approximately 011.747° (rounded to three decimal places).
To summarize:
- The bearing of A from C is approximately 011.747°.
- The distance BC is approximately 59.2075m.
If A is at (0,0), then
B is at (285sin78°,285cos78°) = (278.77,59.25)
Now we know that C is at (278.77,59.25-c) where
278.77^2 + (59.25-c)^2 = 307^2
c = 187.84
So, C is at (278.77,-128.59)
So, in triangle ABC, using the law of cosines,
285^2 = 187.84^2 + 307^2 - 2*187.84*307*cosA
A = 65°
Thus the bearing of A from C is 360-A = 295°
As noted above, BC = 187.84