Are the lengths better now. I changed the 5 to a 11. I was then going to apply cosine but I’m getting a negative . Helppp pls
Probably applying the cosine wrong then. In fact I'm not even sure why you need to use cosine since you label the opposite and adjacent sides, and the hypotenuse is extra work, but OK you do you.
Probably applying the cosine wrong then. In fact I'm not even sure why you need to use cosine since you label the opposite and adjacent sides, and the hypotenuse is extra work, but OK you do you.
Also you carried your error of sqrt(41) forward from your previous mistake, AND back there you calculating this on the basis that the triangle is right-angled (hence Pythag) so I'm really not understanding your logical car crash here.
Are the lengths better now. I changed the 5 to a 11. I was then going to apply cosine but I’m getting a negative . Helppp pls
You have corrected the horizontal distance from 5 to 11, which is fine, but you have left the hypotenuse unchanged. But as @RDKGames has said, you don't even need the hypotenuse. The tangent of the marked angle at A is 11/4. This is the angle between South and AB. The bearing of B from A is the angle between North and AB.
You have corrected the horizontal distance from 5 to 11, which is fine, but you have left the hypotenuse unchanged. But as @RDKGames has said, you don't even need the hypotenuse. The tangent of the marked angle at A is 11/4. This is the angle between South and AB. The bearing of B from A is the angle between North and AB.
Town. A. is 3km west and 6km north of fixed point o. Town B is 8km east and 2km north of foxed point o. Find the bearing of B to A?
I just don't see how to get the answer to be 110
Point A & B make a right angle triangle with a base length of 11 and height of 4, the angle at point A can be found using tan and then subtracting that answer from 180 will give you the bearing.