Question

two forces of 430 newtons and 425 newtons act on a point. the resultant force is 826 newtons. find the angle between the two forces.
the angle between the two forces is (square^{circ}).
(round to the nearest integer as needed.)

Explanation:

Step1: Recall the Law of Cosines for forces

The formula for the magnitude of the resultant force \( R \) of two forces \( F_1 \) and \( F_2 \) with an angle \( \theta \) between them is \( R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos\theta \) (Wait, no, the correct Law of Cosines for the magnitude of the resultant when two vectors (forces) are added is \( R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos\theta \)? Wait, no, actually, the angle between the two vectors is \( \theta \), and the formula for the magnitude of the resultant \( R \) is \( R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos\theta \)? Wait, no, let's correct. The Law of Cosines for vector addition: if we have two vectors \( \vec{F_1} \) and \( \vec{F_2} \) with an angle \( \theta \) between them, then the magnitude of the resultant \( \vec{R}=\vec{F_1}+\vec{F_2} \) is given by \( R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos\theta \)? Wait, no, actually, the angle in the Law of Cosines for the triangle formed by the vectors is \( 180^{\circ}-\theta \) if \( \theta \) is the angle between the two vectors when placed tail - to - tail. Wait, let's start over.

Let \( F_1 = 430\) N, \( F_2=425\) N, and \( R = 826\) N. The formula for the magnitude of the resultant of two vectors is \( R^{2}=F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\cos\theta \), where \( \theta \) is the angle between the two vectors (when they are placed tail - to - tail).

So we can solve for \( \cos\theta \):

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Step2: Substitute the values

First, calculate \( R^{2}\), \( F_{1}^{2}\), and \( F_{2}^{2}\):

\( R^{2}=826^{2}=826\times826 = 682276\)

\( F_{1}^{2}=430^{2}=430\times430 = 184900\)

\( F_{2}^{2}=425^{2}=425\times425 = 180625\)

Then, substitute into the formula for \( \cos\theta \):

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Step3: Find the angle \( \theta \)

Now, we find \( \theta=\arccos(0.8666) \)

Using a calculator, \( \arccos(0.8666)\approx30^{\circ} \) (since \( \cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.8660 \), and our value 0.8666 is close to that, and when we calculate \( \arccos(0.8666) \) using a calculator, we get approximately \( 30^{\circ} \) (rounded to the nearest integer))

Answer:

\( 30 \)