A rope extends between 2 poles.  A 90n body hangs from it.  Find the
tension in the 2 parts of the rope.

y = FT1sin10degrees + FT2sin5degress-90n = 0
FT1y = FT1sin10degrees
FT2y = FT2sin5degrees

X = FT2cos5degrees - FT1cos10degrees = 0
FT1FT2 = 343N
       = 346N

I don't understand how the 346N figure was reached.

Hello again.

There should be a diagram with your question that shows the 10 degree
and 5 degree angles.  I can't draw it here either, but just picture
the 90 N weight hanging a bit off center and pulling the horizontal
rope down only a little bit, so there's a 5 degree angle on one side
and a 10 degree angle on the other.

In the equation
y = FT1sin10degrees + FT2sin5degress-90n = 0

they're telling you that the downward force of gravity pulling on the
90N weight is being balanced by the vertical component of each of the
two rope segments that support it.

The tension in each of the two rope segments is the length of the
hypotenuse of the triangle that forms when they sag down.  The
vertical leg of each triangle is the Y component of that tension
force, and the horizontal leg of each triangle is the X component.

Since the wait is motionless and hangs straight down, the two Y
components (pulling up) add up to the 90N that's pulling straight
down.
And the two X components (pulling sideways in opposite directions) add up to zero. 

The two vertical components are the first hypotenuse * sin 10 degrees
plus the second hypotenuse * sin 5 degrees.  

y = FT1sin10degrees + FT2sin5degress-90n = 0
means that the two of them minus 90N add up to zero
(I'd rather say 
FT1sin10degrees + FT2sin5degress  =  90n
but that's the same thing)

The two horizontal components are the first hypotenuse * cos 10 degrees
the second hypotenuse * cos 5 degrees
and since they are pulling opposite ways and add up to zero you can write
X = FT2cos5degrees - FT1cos10degrees = 0
(I'd rather say 
FT2cos5degrees = FT1cos10degrees 
but that's the same thing)

Now it's just algebra.  You have two equations and two unknowns
So from the second equation
FT2 = FT1cos10degrees / cos5degrees
    = FT1 * 0.984807753 / 0.996194698
    = FT1 * 0.988569559
Now you go back to the first equation

FT1 sin10degrees + FT1 * 0.988569559 * sin5degrees  =  90
FT1 0.173648178  + FT1 * 0.988569559 * 0.087155743  =  90
0.259807692  FT1 = 90
FT1 = 346N
and plugging back in above,
FT2 = 343N

No search terms for this one--I knew how to do it!

Sincerely,
Richard-ga

Hello Richard,
Thank you in responding to my question.  Your explanation was very
clear, easy to understand.  Really helped.  Thank you again.  I work
part-time, so financial situation at the moment is somewhat bleak,
this tip is for both questions you answered.
Maria
Maria

Maria

Thank you so much for the kind words and your generous tip

-Richard