As was foretold, we've added advertisements to the forums! If you have questions, or if you encounter any bugs, please visit this thread: https://forums.penny-arcade.com/discussion/240191/forum-advertisement-faq-and-reports-thread/

SoonerMan
Registered User

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.

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.

Rah, Oklahoma! Rah, Oklahoma! Rah, Oklahoma~! O-K-U!

0

## Posts

clsCorwinonSee how many books I've read so far in 2010

SerpentonSoonerManonAngle 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.

WerewulfyonStart 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.

clsCorwinonSee how many books I've read so far in 2010

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.

SoonerManonsqrt((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.

redxonSoonerManonyou 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.

redxon