two forces act on a body so the resultant is a force of 68.3 pounds. the angles between the resultant and…

two forces act on a body so the resultant is a force of 68.3 pounds. the angles between the resultant and the forces measure 46 degrees and 38 degrees. the applied forces have different magnitudes. what is the larger magnitude of these forces? 49.4 pounds 52.7 pounds 41.5 pounds 53.6 pounds

two forces act on a body so the resultant is a force of 68.3 pounds. the angles between the resultant and the forces measure 46 degrees and 38 degrees. the applied forces have different magnitudes. what is the larger magnitude of these forces? 49.4 pounds 52.7 pounds 41.5 pounds 53.6 pounds

Answer

Explanation:

Step1: Use the law of sines

Let the magnitudes of the two forces be $F_1$ and $F_2$, and the resultant force $R = 68.3$ pounds. The angles between the resultant and the forces are $\alpha=46^{\circ}$ and $\beta = 38^{\circ}$. The third - angle of the triangle formed by the two forces and the resultant is $\gamma=180^{\circ}-(46^{\circ}+38^{\circ})=96^{\circ}$. By the law of sines, $\frac{F_1}{\sin\alpha}=\frac{F_2}{\sin\beta}=\frac{R}{\sin\gamma}$.

Step2: Calculate the two forces

We know that $\frac{F_1}{\sin46^{\circ}}=\frac{F_2}{\sin38^{\circ}}=\frac{68.3}{\sin96^{\circ}}$. First, find $F_1$: $F_1=\frac{68.3\sin46^{\circ}}{\sin96^{\circ}}$. Since $\sin46^{\circ}\approx0.7193$ and $\sin96^{\circ}\approx0.9945$, then $F_1=\frac{68.3\times0.7193}{0.9945}\approx49.4$ pounds. Second, find $F_2$: $F_2=\frac{68.3\sin38^{\circ}}{\sin96^{\circ}}$. Since $\sin38^{\circ}\approx0.6157$ and $\sin96^{\circ}\approx0.9945$, then $F_2=\frac{68.3\times0.6157}{0.9945}\approx41.5$ pounds.

Answer:

49.4 pounds