8 $$ 1 \leftarrow $$ Question 6 of 25 , Torces of 412 N and 592 N act on an object. The angle between the forces is $$ 46^{\circ} $$. Find the magnitude of the resultant and the angle that it makes with the larger force. 25 point(s) possible The magnitude of the resultant is approximately $$ \square \mathrm{N} $$. (Round to the nearest whole number as needed.) The angle that the resultant makes with the larger force is $$ \square^{\circ} $$. (Round to the nearest degree as needed.)

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The two forces are  
$$
F_1 = 412\text{ N}, \quad F_2 = 592\text{ N},
$$  
with an angle between them of  
$$
\theta = 46^\circ.
$$

**Step 1. Compute the Magnitude of the Resultant**

Using the cosine law for vector addition, the magnitude $$ R $$ of the resultant is given by  
$$
R = \sqrt{F_2^2 + F_1^2 + 2 F_2 F_1 \cos\theta}.
$$

Substitute the given values:
$$
R = \sqrt{592^2 + 412^2 + 2 \cdot 592 \cdot 412 \cos(46^\circ)}.
$$

Calculate the squares:
$$
592^2 = 350{,}464, \quad 412^2 = 169{,}744.
$$
So,
$$
F_2^2 + F_1^2 = 350{,}464 + 169{,}744 = 520{,}208.
$$

Next, compute the product term:
$$
2 \cdot 592 \cdot 412 = 487{,}808.
$$
Using $$\cos(46^\circ) \approx 0.6946$$,
$$
2 F_2 F_1 \cos(46^\circ) \approx 487{,}808 \times 0.6946 \approx 338{,}832.
$$

Now add these two contributions:
$$
R^2 \approx 520{,}208 + 338{,}832 = 859{,}040.
$$

Taking the square root gives:
$$
R \approx \sqrt{859{,}040} \approx 927\text{ N}.
$$

**Step 2. Compute the Angle the Resultant Makes with the Larger Force**

Let the larger force $$ F_2 = 592 \text{ N} $$ be along the horizontal axis. The angle $$ \alpha $$ that the resultant makes with $$ F_2 $$ is given by  
$$
\alpha = \tan^{-1} \left(\frac{F_1 \sin\theta}{F_2 + F_1 \cos\theta}\right).
$$

Substitute the given values and use:
$$
\sin(46^\circ) \approx 0.7193, \quad \cos(46^\circ) \approx 0.6946.
$$

Compute the numerator:
$$
F_1 \sin(46^\circ) \approx 412 \times 0.7193 \approx 296.5.
$$

Compute the denominator:
$$
F_2 + F_1 \cos(46^\circ) \approx 592 + 412 \times 0.6946 \approx 592 + 286.1 \approx 878.1.
$$

Thus,
$$
\alpha \approx \tan^{-1}\left(\frac{296.5}{878.1}\right) \approx \tan^{-1}(0.3376) \approx 19^\circ.
$$

**Final Answers:**
- The magnitude of the resultant is approximately $$927 \text{ N}$$.
- The angle that the resultant makes with the larger force is approximately $$19^\circ$$.
The magnitude of the resultant force is approximately 927 N, and it makes an angle of about 19 degrees with the larger force.

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