Is it from PEP? |F 1|=40N, making an angle θ with the horizontal plane; |F2|=70N, along the horizontal direction, |F|= 100N of the two resultant forces, find the included angle θ and β between f and the horizontal line. . .

The problem solving process is as follows:

(1) According to the parallelogram rule, vector F 1+ vector F2= vector f (next, directly omit two lines of the vector? )

Then square both sides at the same time to get F 1? +2F 1F2+F2？ =F？ = 10000N

F 1？ = 1600，F2？ =4900

∴2f 1f2=3500,f 1f2= 1750

And ∵ f1F2 = | f1| F2 | * cosθ = 40 * 70 * cosθ = 2800cosθ =1750.

∴cosθ= 1750/2800=5/8。 The first problem is solved. By the way, there is no need to ask for the exact value of θ.

(2) ∵ vector F 1+ vector F2= vector f.

∴ vector F 1= vector f- vector F2

Then both sides are squared at the same time,

Get F 1? =F？ -2F*F2+F2？ =F？ -2|F|*|F2|*cosβ+F2？ = 10000-2 * 100 * 70 * cosβ+4900 = f 1？ = 1600

Simplified to-14000cosβ =1600-14900 =-13300.

∴cosβ= 13300/ 14000= 19/20

It's done. Well, I'm a freshman now. I learned this last semester, and I almost forgot. . . .

I'm afraid you don't understand, so the process is a bit verbose. Never mind. Ask me if you don't understand anything!