D = 5 + 8i
Tan A = x/y = 5/8.
A =
A student travels 8km north and then 5km east. What is then her bearing from her starting point?
6 answers
The student travel 8km north and 5km east .What is the bearing of here starting point?
2+2
032degree
D=√(8^2+5^2)
√(64+25)=
√89=9.43m
√(64+25)=
√89=9.43m
To find the bearing of the starting point, we need to use trigonometry. We can use the tangent function to find the angle that the student traveled from her starting point.
Let A be the angle between the student's path and due north. Then:
tan(A) = opposite/adjacent = 5/8
A = tan^-1(5/8) ≈ 32.05 degrees
So the bearing from her starting point is 32.05 degrees east of due north.
Let A be the angle between the student's path and due north. Then:
tan(A) = opposite/adjacent = 5/8
A = tan^-1(5/8) ≈ 32.05 degrees
So the bearing from her starting point is 32.05 degrees east of due north.
1 answer