Stop playing around amd start paying attention in class STUPID KIDS
We know that
<span>The measure of the external angle is the semi-difference of the arcs that it covers.
</span>so
m∠BCD=(1/2)*[arc AE-arc BD]
20=(1/2)*[85-BD]
40=85-BD
BD=85-40
BD=45°
the answer is
BD=45°
This is a vector problem so we will split it by components and add them.
x direction
200sin(240) + 60sin(25) = -147.847985
y direction
200cos(240) + 60cos(25) = -45.6215327
To find the actual speed we now find the magnitude
(-147.847985)^2 + (-45.6215327)^2 = c^2
c = 154.72669748 mph
To find the direction we can do arctan of the different directions:
arctan(-147.847985/-45.6215327) = 252.85 degrees.
So the plane is going 154.73 mph at 252.85 degrees :)
