ok i created a right angle triangle...
tan(theta) = opp /adj
hypotenuse is 7.89
rx = 7.36
ry = 2.87
tan theta = 7.89 / 7.36
= 46.99 degrees
Does this look right? :nb)
Yes now I do. Phew I went a long way from my original thread, I appreciate the help Chestermiller.
So for part c,
R = rx^2 + ry^2
= (7.36)^2 + (2.87)^2
=7.89
As for part d,
I would have to use the sine law? I am totally stuck here.
R = 7.89
angle = 23 degrees
B = 8.0
angle = ?
angle R...
magnitude of R on the x axis:
ax = 6cos90 = 0
bx = 8cos23 = 7.36
rx = ax + bx
rx = 7.36
along the y axis:
ay = 6sin90 = 6
by = -8sin23 = - 3.13
ry = ay + by
ry = 2.87
B is pointing down, meaning its negative... but I still don't know what I'm supposed to do with that information .. Am i...
Hmm I'm not quite sure how to do that... Do I have to solve for the x and y-axis separately?
for x-axis do i have to do this:
8cos23 =
6cos23 =
and y axis:
8sin23 =
6sin23 =
And I'm also confusing myself because am I supposed to use 23 degrees or 67degrees? and why?
Hi everyone,
I need steps and some guidance on how to solve questions like this. I know how to do right triangle questions using the Pythagorean method, but when it comes to these non right angle triangles I get so confused. Help!
1. Homework Statement
Homework Equations
a) The magnitude...