Trigonometry Problem

SoonerMan

I can't figure out my girlfriend's homework. Here is a crude image in paint of what it looks like.

angle I is 15 degrees.
We figure that both angles II and III are 75 and 105 degrees respectively because of the right angle formed by the pole atop the slope. We are trying to find how long X1 and X2 are.

The image is a pole on a slope with two wires holding it down, we're trying to figure out how long the wires are. The problem is that we cannot figure out how we can find the other angles. I think that the 15 degrees comes into play again, but I can't figure out how.

reply ASAP, please and thanks.

clsCorwin

Make that picture bigger, cuz I can't figure out whats going on with it that small, FYI.

Serpent

you are not giving all the information. are there any lengths given?

SoonerMan

There you go, size and lengths given. Sorry. Please give a step by step instruction on how to get the other angles. She's supposed to solve it by Law of Sin if that helps.

Werewulfy

You'll actually be using the Law of Cosines for this one, or at least, I don't know how to use Law of Sines for it.

Angle One will be 180 - 90(assuming that bottom line is a straight one) - 15 = 75
Angle Two 180 - 75 = 105, since Angle One and Two should add up to be 180.

Now the angles in the triangles (right above one and two) will be the opposite, once again, 180 - angle one = 105, and 180 - angle two = 75. Then you just plug in the angle and sides into this equation.

c^2 = a^2 + b^2 - 2ab*cos(C) or

x1^2 = 30^2 + 60^2 - 2(30)(60)*cos(105) x1 = 73.7

x2^2 = 30^2 + 60^2 - 2(30)(60)*cos(75) x2 = 59.2


I think that should be right.

clsCorwin

(sin A / a) = (sin B / b) = (sin C / c) is the law of sines, if I remember correctly.

Start by labeling the triangle appropriately, sides a, b, c and angles A, B, C respectively.

Your deductions about angles 1 and 2 are correctly, and also give insight into the triangle. The angle above 1 is going to be equal to angle 2, and vice versa.

Lets call your angle 1, C, and angle 2, F. sin C / x1 = sin B / 30.

B is going to be the angle on the bottom left. Likewise, call the opposite side E, and use sin F / x2 = sin E / 30.

You just need to find that angle. I gotta run, so I'll let someone else fill in the gaps if I don't get back in time.

SoonerMan

Thanks so much. The past few problems on her review were using Law of Sin so I was kind of stuck on that. Now I'm horrible with bearings and she can't figure this one out either.

A plane flies 250 miles bearing 218 degrees then turns to bearing 140 degrees and flies 150 miles. How far from the start?

This one I'm assuming after finding the angle of turn you use law of sin, but I could be wrong. From the starting point we believe the angle is 38 (218-180) and we get part of the second angle to be 50 degrees. Thanks again in advance.

redx

uhh... well pretty sure bearing is just the direction the plane is heading. you can just assume that it is heading north, or whatever you want. Then the plan turn 78 degrees and flys some more.

sqrt((250 + (cos 78)150)^2 + (150 (sin 78))^2)

or something like that. Like, it only turns once, and amount that it turns is all that matters (218-140=78) cause you are just looking for distance.


I think.

it is like, 317 or so... I doubt you would be working with degrees if they expected non estimated/rounded values.

SoonerMan

Thanks, but the fact that the bearing is 240 degrees means it's going southwest, and when it turns to the bearing of 140 it is now going southeast. So where the bearings are is crucial to the angles.

redx

naw... just draw it out.

you get a line 250 miles long south west, and one 150 south east.

now, just turn the paper. The total distance does not change.

so... just spin it about so 250 is along the x axis. 250 plus the x component of the 150 line is the total x displacement. the y component of the 150 line is the total y displacement. The hypotenuse is the total displacement.