Two boats A and B left port C at the same time on different routes B traveled on a bearing of 150° and A traveled on the north side of B.when A had traveled 8km and B had traveled 10km. The distance between the two boats was found to be 12km.calculate the bearing of A's route from C.

Question

Two boats A and B left port C at the same time on different routes B traveled on a bearing of 150° and A traveled on the north side of B.when A had traveled 8km and B had traveled 10km. The distance between the two boats was found to be 12km.calculate the bearing of A's route from C.

Answer 1

You travel on a heading, not a bearing. Math texts drive me crazy.

triangle ABC has sides 8, 10,12
lets find angle ACB
12^2 = 10^2 + 8^2 - 2 * 10 *8 cos C
160 cos C = -144 + 100+ 64 = 20
cos C = 20/160 = .125
C = 82.8 degrees
so
Angle of AC clockwise from north = 150 - 82.8 = 67.2 degrees about East North East

Answer 2

To calculate the bearing of A's route from port C, we need to understand the relative positions and distances traveled by both boats A and B.

First, let's draw a diagram to visualize the situation:

```
B (10km)
^
/
/
/
/
/ 12km
/
/
/
/ θ (unknown bearing of A's route)
/
/
C ----------> A (8km)
```

In this diagram, C represents port C, A represents boat A, B represents boat B, and θ represents the unknown bearing of A's route from C.

Now, let's break down the information we have:

- Boat B traveled on a bearing of 150°.
- Boat A traveled on the north side of B.
- When A had traveled 8km and B had traveled 10km, the distance between the two boats was found to be 12km.

To find the bearing θ, we can use the law of cosines. This law relates the lengths of the sides of a triangle to the cosine of one of its angles.

In this case, we have a triangle formed by points C, A, and B, with sides of lengths 8km, 10km, and 12km, respectively. Let's label the length of side AB as c (12km), the length of side AC as a (8km), and the length of side BC as b (10km).

The law of cosines states that:

c^2 = a^2 + b^2 - 2ab * cos(C)

In our case, C is the angle at point C. We want to find the angle θ, which is the complement of angle C.

Rearranging the equation, we get:

cos(C) = (a^2 + b^2 - c^2) / (2ab)

Using the given values, we can substitute them into the equation:

cos(C) = (8^2 + 10^2 - 12^2) / (2 * 8 * 10)

cos(C) = (64 + 100 - 144) / 160

cos(C) = 20 / 160

cos(C) = 0.125

To find the value of C, we can take the inverse cosine (arccos) of 0.125:

C = arccos(0.125)

C ≈ 82.02°

Since the bearing θ is the complement of angle C, we can calculate it as:

θ = 180° - C

θ ≈ 180° - 82.02°

θ ≈ 97.98°

Therefore, the bearing of A's route from port C is approximately 97.98°.