City A is 300km due east of city B.city C is 200km on an bearing of 123degree from city B how far is it from C.
Did you mean, how far is B from A?
You just said it is 200 km from C.
Make a sketch
I have triangle ABC, with BC = 200, CA = 300 and angle C = 147°
Use the cosine law to find BA
All angles are measured CW from +Y-axis.
BA = 300km[90o].
BC = 200km[123o].
CA = 300km[90o] - 200km[123o].
X = 300*sin90 - 200*sin123 = 132.3 km.
Y = 300*Cos90 - 200*Cos123 = 109 km.
CA = sqrt(X^2 + Y^2).
y
c
To find the distance from city C to city B, we can use trigonometry.
First, let's find the horizontal and vertical components of the displacement from city B to city C.
The horizontal component can be found using the formula:
Horizontal Component = Distance * cosine(angle)
Given that the distance is 200 km and the angle is 123 degrees, we can calculate the horizontal component as follows:
Horizontal Component = 200 km * cosine(123 degrees)
Next, let's find the vertical component using the formula:
Vertical Component = Distance * sine(angle)
Given that the distance is 200 km and the angle is 123 degrees, we can calculate the vertical component as follows:
Vertical Component = 200 km * sine(123 degrees)
Now, we can use these components to find the distance from city C to city B. We can use the Pythagorean theorem, which states that the square of the hypotenuse (distance) is equal to the sum of the squares of the other two sides (horizontal and vertical components).
Distance^2 = Horizontal Component^2 + Vertical Component^2
Plugging in the values we calculated earlier, we get:
Distance^2 = (200 km * cosine(123 degrees))^2 + (200 km * sine(123 degrees))^2
To find the distance, we take the square root of both sides:
Distance = sqrt((200 km * cosine(123 degrees))^2 + (200 km * sine(123 degrees))^2)
Evaluating this expression will give us the distance from city C to city B.