A 90 N boy is hanging on a rope that extends between two poles. Find the tension of the 2 ropes. The angles were 5 and 10 degrees

Answer 1

Your diagram probably shows more than what you stated,
e.g. are the 2 points of contact of the ropes at the same horizontal plane? I will assume they are.

Let the tension in the rope at 5° be t1
let the tension in the rope at 10° be t2

for the horizontal vectors:
t1 cos5 = t2 cos10 -----> t1 = t2 cos10/cos5 ***

for the vertical vectors:
t1 sin5 + t2 sin10 = 90
using substitution:
(t2cos10/cos5)(sin5) + t2 sin10 = 90
mutiply each term by cos5
t2 cos10 sin5 + t2 sin10 cos5 = 90cos5
t2(sin5cos10 + cos5sin10) = 90cos5
t2( sin15) = 90 cos5
t2 = 90 cos5/sin15 = ....

check my work, I did not write it out in full first.
Btw, I was using the identity :
sin(A+B) = sinAcosB + cosAsinB in the 2nd last step above to get sin 15°

once you have that , you can find t1 from ***

Sal Khan has a video of this type of problem.
It is one of his earlier ones, so it might be hard to read some of the diagrams.

https://www.khanacademy.org/science/physics/forces-newtons-laws/tension-tutorial/v/tension-part-2

Answer 2

THANK YOU REINY :))