1. ## Angles

Ok - So 2 ends of a cord of length 5.6m are attached to a beam 4.8m apart. A shoe is hanging off the cord 2.1m from one end.

When i'm calculating the angles that are the 2 sections of cord meet the beam, is this correct...

Using the cosine rule, we can asign a = 5.6, b = 4.8, c = 2.1

So...

$CosA = \frac{4.8^2 + 2.1^2 - 5.6^2}{2 * 4.8 * 2.1} = -0.193948$

So $arccos(-0.193948) = 101.18$

I'd then do the same to find the other angle?

Ok - So we have a cord in length 5.6m, which are attached to a length 4.8m apart. A shoe is hanging off the rope 2.1m from one end.

When i'm calculating the angles that are the 2 sections of cord meet the length, is this correct...

Using the cosine rule, we can asign a = 5.6, b = 4.8, c = 2.1

So...

$CosA = \frac{4.8^2 + 2.1^2 - 5.6^2}{2 * 4.8 * 2.1} = -0.193948$

So $arccos(-0.193948) = 101.18$

I'd then do the same to find the other angle?

correct then you can use sine low for less calculations

$\frac{a}{\sin A} = \frac{b}{\sin B}$

3. But surely i'll end up with 3 angles and i only need two?

so where do i go from here?

But surely i'll end up with 3 angles and i only need two?

so where do i go from here?
I solved it, since I think we have a triangle

the question said a chord 5.6m attached to a another one with length 4.8 right ??
then a shoe is hanging off the rope 2.1 m from one end

the end of what ?? and the shoe is hanging off which rope you said before chord then rope ??

can you explain

5. I make 3 angles:

101.8 , 57.23, 21.58

but i dont understand which two angles relate to the rope coming away from the beam

6. after the outcome to the above i need to draw a force diagram and then a triangle of forces where the shoe weighs 36g..any ideas?

Ok - So 2 ends of a cord of length 5.6m are attached to a beam 4.8m apart. A shoe is hanging off the cord 2.1m from one end.

finally I get it this is the graph, okay all things are clear what the question want can you tell me

8. Thats the one

So trying to figure out the angles that meet the 4.8 line...

9. call the angle between 2.1 and 3.5 A so the side 4.8=a the side 2.1=b and the side 3.5=c

as you solved before by cosine low

$b^2 = a^2 + c^2 - 2ac\cdot \cos B$

$(2.1)^2 = (4.8)^2 + (3.5)^2 - 2(3.5)(4.8) \cos B$

$\cos B = \frac{4.8^2 + 3.5^2 - 2.1^2 }{2(3.5)(4.8)}$

$B = \cos ^{-1} \left(\frac{4.8^2 + 3.5^2 - 2.1^2 }{2(3.5)(4.8)}\right)$

you can continue

10. So we're saying b = 23.07

So do we just repeat that same process for c ?

So we're saying b = 23.07

So do we just repeat that same process for c ?

ya or you can use sine law

12. so C = 40.78 ?

13. and therefore A = 117.15?

please could you confirm my results

14. How can this relate to a forces diagram?