City y is 200km and in the direction of 048 from city x. City x is 80km and in the direction of 108 from city y. Find the distance of city z from city x and the bearing of city
2 answers
City y is 200km and in the direction of 048 from city x. City x is 80km and in the direction of 108 from city y. Find the distance of city z from city x and the bearing of city x from z.
You repeated your post with the same error in both.
I will assume that your second sentence should say:
City z is 80km and in the direction of 108 from city y
I sketched the diagram and got triangle xyz to have an angle of 120° at y
So by the cosine law:
d^2 = 200^2 + 80^2 - 2(200)(80)cos120°
find d
once you have d, we can use the sine law to find angle yxz and you can get
the bearing easily after that. (add it to 48°)
I will assume that your second sentence should say:
City z is 80km and in the direction of 108 from city y
I sketched the diagram and got triangle xyz to have an angle of 120° at y
So by the cosine law:
d^2 = 200^2 + 80^2 - 2(200)(80)cos120°
find d
once you have d, we can use the sine law to find angle yxz and you can get
the bearing easily after that. (add it to 48°)
1 answer