You walk $53$ m to the north, then turn $60$ degrees to your right and walk another $45$ m. Determine the direction of your displacement vector. Express your answer as an angle relative to east.
So I did pythagorean theorem and got the adjacent side to be $28$. Then I did inverse tan and got the angle to be $50$ degrees. Help please?
The answer is $63$ degrees.
On
If you were to draw a graph, the first leg of the trip would go straight up, and once you reach the second leg of the trip you can recognize that a triangle forms above it, looking like a flag pole with a triangular flag at the top. Using your properties of that smaller triangle, you can now look at the bigger triangle and solve from there.
From here you can find the total height of the larger triangle to be $53+45\cdot \frac{1}{2} = 75.5$ and the width to be $\frac{45\sqrt{3}}{2}$.
Using Inverse tangent on these will give you the angle at the bottom corner of the triangle. Since it asks for angle relative to east however, you'll want to take 90 degrees minus that to get the complement.
If we are starting from point $(0,0)$ then after the first step we are in $(0,53)$ and after the second we got to $(0 + 45\sin \alpha, 53 + 45 \cos \alpha)$. $\sin \alpha = \frac{\sqrt 3}{2}, \cos \alpha = \frac 12$, so the resulting point is $(\frac{45\sqrt3}{2},\frac{151}{2})$ and, as our starting point was $(0,0)$, this is also the displacement vector.