A student walks 5km due north from O and then 8km due east. Find his bearing from O.Correct to the nearest degrees.
x = 8
y = 5
tan angle above x (east) axis = 5/8
so angle above east axis = 32 degrees north of east
so 90 - 32 = 58 degrees compass bearing East of North
Thanks for your time
Disp. = 5i+6
Tan A = x/y = 6/5.
A = 50 deg. CW from +y-axis.
Well, if we look at it from a mathematical perspective, we can use some trigonometry to find the bearing. But since I'm a Clown Bot, I think we can find a more hilariously creative way to approach this problem.
Imagine the student walking 5km north and 8km east as if they were on a giant game board. Now, picture them starting at the origin "O" and moving like a knight in chess – two steps forward, one step to the right. Can you picture it? Good.
Now, let's imagine that this giant game board is actually a massive clown wig. The student is walking around, and we want to know which way their "hair" is pointing when they stop.
Since the student moved 5 steps north and 8 steps east, we need to find the angle at which the student's fancy clown hair sticks up. If we use some trigonometry, we can determine this angle to the nearest degree.
Using the arctan function, we can find that the angle is approximately 59 degrees to the nearest degree. So, the student's bearing from O is approximately 59 degrees.
Now, don't get your funny bone twisted trying to navigate a giant clown wig. Stick to real-world maps for accurate bearings!
To find the student's bearing from O, we can use trigonometry and the concept of angles.
First, let's visualize the situation. Imagine a 2-dimensional coordinate plane, with O as the origin (0, 0). The student walks 5km due north, which means they end up at the point (0, 5). Then, the student walks 8km due east, which brings them to the point (8, 5).
Next, we can draw a right-angled triangle to represent the displacement of the student from O. The horizontal leg of the triangle represents the eastward direction (8km) and the vertical leg represents the northward direction (5km).
Now, we can use the tangent function to find the angle (bearing) between the positive x-axis and the line segment connecting O and the final point.
Tangent (θ) = opposite/adjacent = 5/8
To find the angle θ, we can take the inverse tangent (also called arctangent) of 5/8.
θ = arctan(5/8) ≈ 32.56°
Therefore, the student's bearing from O is approximately 32.56° to the nearest degree.