An object is suspended from two cables that meet at a point above the object. One cable pulls with a force modeled by F1 = −77i + 35j. The other cable pulls with a force modeled by F2 = 92i + 84j. What is the angle between the cables? Round to the nearest degree.

[tex]\begin{gathered} F_1\cdot F_2=|F_1|\cdot|F_2|\cdot\cos \theta \\ \Rightarrow\cos \theta=\frac{F_1\cdot F_2}{|F_1|\cdot|F_2|} \end{gathered}[/tex]

where θ = angle between the forces

Let us find the dot product of the two forces:

[tex]\begin{gathered} F_1\cdot F_2=(-77i+35j)\cdot(92i+84j) \\ F_1\cdot F_2=(-77\cdot92)+(35\cdot84) \\ F_1\cdot F_2=-7084+2940 \\ F_1\cdot F_2=-4144 \end{gathered}[/tex]

Now, let us find the magnitude of the two vectors.

For F1:

[tex]\begin{gathered} \lvert F_1\rvert=\sqrt[]{(-77)^2+(35)^2} \\ \lvert F_1\rvert=\sqrt[]{5929+1225}=\sqrt[]{7154} \\ \lvert F_1\rvert=84.58 \end{gathered}[/tex]

For F2:

[tex]\begin{gathered} \lvert F_2\rvert=\sqrt[]{(92)^2+(84)^2} \\ \lvert F_2\rvert=\sqrt[]{8464+7056}=\sqrt[]{15520} \\ \lvert F_2\rvert=124.58 \end{gathered}[/tex]

Now, substitute all the calculated values into the dot product equation obtained above:

[tex]\begin{gathered} \cos \theta=\frac{-4144}{84.58\cdot124.58} \\ \cos \theta=\frac{-4144}{10536.98}=-0.3933 \end{gathered}[/tex]

Solve for θ:

[tex]\begin{gathered} \theta=\cos ^{-1}(-0.3933) \\ \theta=113\degree \end{gathered}[/tex]

That is the angle between the forces.