Arocket fires two engines simultaneously. one produces a thrust of 725 n
directly forward, while the other gives a 513-n thrust at 32.4° above the forward
direction. find the magnitude and direction of the resultant force acting on the rocket.

First resolve the 513N thrust into two components of the right angles.

In the forward side, we have 513N cos(32.4°)

At right angles, we have 513N sin(32.4°)

So...

513N cos(32.4°) = 433.14 N

513N sin(32.4°) = 274.88 N

Next get the total forward force of the rocket.

725 N + 433.14 N = 1,158.14 N

And the total force at right angles:

0 + 274.88 N = 274.88 N

Next solve the Resultant Magnitude (F) through Pythagorean theorem. 

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F² = 1,341,288.26 + 75,559.01 

F² = 1,416,847.27

F = √1,416,847.27

F = 1,190.3139

F = 1,190.31 

Now that we have resultant magnitude, find the direction by dividing total force exerted at right angles by the total force exerted at the forward side. 

tanC = 274.88 N / 1,158.14 N

tanC = 0.237346089419241

C = tan⁻¹ 0.237346089419241

C = 13.35°

ill do it for 100, also this isnt 80 its 40

god i hope this !