I've tried distributing etc but cannot get the correct answer of 60.1N

 

My book doesn't give the steps.

 

Tsin37 + (1.33T)(sin53) - 100N =0

 

I never know what to do with the (1.33T)(sin53)

 

Any assistance would really be appreciated.

is it [math] T*\sin{37}+1.33*T*\sin{53}-100N=0[/math] ? Eventually, can I use 4/3 exept 1,33 ?

Anything you like if you can explain it

Ok, i'll try

First of all, I coversed degrees to radians

[math]T*\sin(37\pi/180)+\frac{4}{3}*T*\sin(53\pi/180)-100N = 0[/math]

[math]T*\sin(37\pi/180)+\frac{4}{3}*T*\sin(53\pi/180)=100N[/math]

[math]T(\sin(37\pi/180)+\frac{4}{3}*\sin(53\pi/180))=100N[/math]

[math]T=\frac{100}{\sin(37\pi/180)+\frac{4}{3}*\sin(53\pi/180)}N[/math]

And aproximate...

 

pq

I made my own calculations, and the result differed from yours, so I checked with Maple, and it confirmed what I got...

 

[math]T={\frac {300N}{3\,\sin \left( 37 \right) +4\,\sin \left( 53

\right) }}[/math]

 

If you need the steps, let me know.

Those are clearly equivialant, the only difference is the RHS is multiplied by [imath]\frac{3}{3}[/imath].

 

I don't see why you would want to convert to radians, all that is doing is introducing another stage where errors are going to build up.

 

[math]T=\frac{100N}{\sin(37)+1.33\sin(53)}[/math]

 

Is fine, but to reduce rounding errors you'll want to reduce the amount of operations that you ultimately perform. If you note that [imath]\sin(37)=\cos(53)[/imath] (probably not a coincidence), then you can get a nice little polynomial in the denominator.

 

[math]T=\frac{100N}{1-\sin^2(53)+1.33\sin(53)}[/math]

 

And this means that you are only evaluating one sine term instead of two.

Well , we haven't learned trigonimetric functions yet so I can't see it there. But I'll take a look sometime...

And that conversation, I don't remember why I did it. Maybe because Maple doesn't take degrees, really don't know.

 

pq

Why, oh why did I miss that [math]4/3[/math] in your result Pq...apologies.