A man start from point A and walks 5km east to B. He then walks 7km on a bearing of 162 degree to reach point C. Find the bearing and distance of C from A

make your sketch, to see a triangle with sides 5 and 7 and the contained angle of 108°

by the cosine law:

AC^2 = 5^2 + 7^2 - 2(5)(7)cos108
= 95.6311...
AC = √95.6311.. = 9.778... (I stored that in memory)

by sine law:
sinØ/7 = sin108/9.778..
sinØ = .68077...
Ø = appr 42.9° -----> E 42.9° S

or, using vectors:

R = (5cos0 , 5sin0) + (7cos -72, 7sin -72)
= (5, 0) + (2.16312 , -6.6273..)
= (7.163... , -6.6273..)

|R| = √(7.163..^2 + (-6.6273..)^2)
= 9.779 , same as above

tanØ = -6.6273../7.163...
Ø = -42.9° ----> E 42.9° S , same as above

To find the bearing and distance of point C from point A, we can use trigonometry and vector addition.

First, let's represent the displacement from A to B as vector AB and the displacement from C to B as vector BC.

Vector AB is 5 km east, so we can represent it as (5, 0) in terms of its components.

To find vector BC, we need to break down the displacement into horizontal and vertical components. The bearing of 162 degrees indicates that the angle is measured clockwise from the positive x-axis.

The horizontal component is BC * cos(162°) and the vertical component is BC * sin(162°). Since the angle is measured clockwise, we use the negative value of sin(162°).

Now, let's calculate the components of vector BC:

Horizontal component of BC = 7 km * cos(162°)
Vertical component of BC = -7 km * sin(162°)

Using a scientific calculator, we can evaluate these values to be approximately:

Horizontal component of BC = -6.601 km
Vertical component of BC = -2.834 km (rounded to three decimal places)

Now, we can add the horizontal and vertical components to find the total displacement from point A to point C:

Horizontal component = 5 km + (-6.601 km) = -1.601 km (rounded to three decimal places)
Vertical component = 0 km + (-2.834 km) = -2.834 km (rounded to three decimal places)

The components of vector AC are approximately (-1.601, -2.834).

Using the Pythagorean theorem, we can find the magnitude (distance) of vector AC:

Magnitude of AC = √((-1.601)^2 + (-2.834)^2) = √(2.562 + 8.049) = √10.611 ≈ 3.259 km (rounded to three decimal places)

The bearing of point C from point A can be found using trigonometry. Let's calculate the angle:

Tan(θ) = vertical component / horizontal component
Tan(θ) = -2.834 km / -1.601 km ≈ 1.770
θ ≈ Tan^(-1)(1.770) ≈ 59.835° (rounded to three decimal places)

However, to determine the bearing from A to C, we need to consider that vector AC is in the second quadrant. Therefore, we need to subtract the calculated angle from 180° to obtain the bearing:

Bearing of C from A = 180° - 59.835° ≈ 120.165° (rounded to three decimal places)

So, the bearing of point C from point A is approximately 120.165° and the distance is approximately 3.259 km.