This is a homwork question and I am trying to figure out part B, can anybody help?

Question

According to the instructions you found with your tent, the Oblique Tent Company has determined that if a support rope on the side of a tent is placed at a 45degree angle to the ground, it will provide maximum support for the tent.

a) How far away from the base of your 6-foot tent wall should you 8ft rope be placed in order to provide maximum support in a strong wind.
This is what I got:
sinC/8=sin45degre/6
sin angle C=70.5 so the 180 - (45+70.5)=64.5.
Sin64.5degree/a=sin45degree/6
The rope should be 7.66 feet away from the base of the tent.

What I don't know how to figure out is the second part:
b) As the tent size changes, so does the length of the wall, the length of the support rope, and the maximum support angle. Derive a general formula that will allow you to determine how far away from the base of the tent wall the end of the support rope should be. Now this is an oblique tent so there is no 90degree angle. Can anybody help figure this out?

Answer 1

When you're given 2 sides and an angle, I find that the law of cosines is usually faster and more accurate.

Let w = the height of the wall, l = the length of the support rope, a = the "maximum support angle", and d = the distance from the base of the wall to the anchor point of the rope.

So since the angle you're given is a, the wall takes the role of c, the side opposite the angle.

w² = l² + d² - 2ldcos(a)

You want a formula for d, so:

d² - (2lcos(a))d + (l² - w²) = 0

That's a quadratic equation in d...very ugly...but you can still technically use the quadratic formula to find d:

d = [2lcos(a) ± √(4l²cos²(a) - 4(l² - w²))]/2

Yes, that's a horrible mess, but to test it, we can see how it compares to your answer for a:

d = [2(8)cos(45) ± √(4(8)²cos²(45) - 4(8² - 6²))]/2
= [16(√2/2) ± √(256(1/2) - 4(28))]/2
= [8√2 ± √16]/2
= [8√2 ± 4]/2
= 4√2 ± 2
= 4√2 + 2 = 7.66 feet, which matches your answer....

OR....(and you missed this in your answer, because for angle C, you could also have gotten 180-70.5 = 109.5 degrees...always have to be careful with law of sines...)

= 4√2 - 2 = 3.66 feet

(This corresponds to the situation where the angle between the tent peg and the ground is obtuse...which doesn't seem too sturdy to me, but then this is just a math problem.)

So as ugly as it is, I think that formula is the best you can do.

Answer 2

According to the instructions you found with your tent, the Oblique Tent Company has determined that if a support rope on the side of a tent is placed at a 45degree angle to the ground, it will provide maximum support for the tent.

The equation is what you use in part a) just use x for the tent wall and y for the rope.

Answer 3

Hi,

You just need to do it like you did part a, but use variables over and over.

In your original triangle, let w = slanted wall length across from the 45 degree angle, X = angle formed between the tent and the ground, Y being the third angle at the top, and z be its opposite side, the distance along the ground from the tent to the place to tie the rope.

If you start out just like you did with your original proportion, you'd have w/sin(45) = r/sin(X)

You know to solve this to get sin(x) = R*sin(45)/w)

To find angle X, you take the inverse sine, so
angle X = sin^-1 (Rsin(45)/w)

Since angle Y is 180 minus the other 2 angles, Y = 180 - 45 - X.
this simplifies that Y = 135 - X. Replacing X with the expression above, you get the formula Y = 135 - [sin^-1 (Rsin(45)/w)]

Then you use the law of sines again, with angles 45 and Y.

sin Y /Z = sin(45)/W

Replacing Y with its expression from above, you get

sin [sin^-1 (Rsin(45)/w)]/Z = sin(45)/W

Cross-multiplying and solving for Z, you get

Z = w*sin [ sin^-1 (Rsin(45)/w)]/ sin(45)

I hope that's clear. Notation loses something online.

Answer 4

actually if you do this graphically , you will find that the length of the rope is a red herring.
the distance from the wall is the same as the height of the wall
then you will get a 45 degree angle.
God bless,
gabe.