Three forces of magnitudes 16, 19, and eight newtons, respectively, are acting at a single point. Given that the forces are in equilibrium, find the measure of the angle between the forces of magnitudes 16 and eight newtons to the nearest minute.
If we consider the three forces acting at the point 𝑝, as they’re in equilibrium, we know that the resultant force is equal to zero. We are trying to calculate the angle between the eight-newton force and the 16-newton force.
If we redraw this diagram so the forces are nose to tail as they’re in equilibrium, we can create a triangle where the angle between the eight-newton force and the 16-newton force is now 180 minus 𝜃. Using the cosine rule, 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴, we can work out the angle in the triangle. From there, we can subtract it from 180 degrees to work out 𝜃, the angle between the 16-newton and eight-newton force.
Substituting in the values from the triangle gives us 19 squared is equal to 16 squared plus eight squared minus two multiplied by 16 multiplied by eight multiplied by cos 𝐴. 19 squared is equal to 361. 16 squared plus eight squared is equal to 320. And two multiplied by 16 multiplied by eight is 256. Therefore, our equation becomes 361 equals 320 minus 256 cos 𝐴.
Rearranging this equation gives us 256 cos 𝐴 is equal to 320 minus 361. And dividing both sides by 256 allows us to work out the value for cos 𝐴. Angle 𝐴 can then be calculated by doing the inverse cos or cos to the minus one of negative 41 divided by 256. This is equal to 99.216 degrees.
As 180 minus 𝜃 is equal to 99.216, we can calculate 𝜃 by subtracting 99.216 from 180. This means that 𝜃 is equal to 80.784 degrees. As the question asked for the angle to the nearest minute, we need to convert 0.784 to minutes. 0.784 multiplied by 60 is 47.04, or 47 to the nearest minute. Therefore, the angle 𝜃 between the 16- and eighteen [eight]-newton forces is 80 degrees and 47 minutes.