# Question Video: Resolving a Force into Two Nonperpendicular Components Mathematics

A force 𝐅 of magnitude 99 N acts due south. It is resolved into two components as shown on the diagram. Find the magnitudes of 𝐅₁ and 𝐅₂ giving values to two decimal places.

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### Video Transcript

A force 𝐅 of magnitude 99 newtons acts due south. It is resolved into two components as shown on the diagram. Find the magnitudes of 𝐅 sub one and 𝐅 sub two, giving values to two decimal places.

In this question, we’re given two components of the force 𝐅, and it is the magnitude of these components 𝐅 sub one and 𝐅 sub two that we need to calculate. One way of doing this is to create a triangle of forces as shown, where the three sides of our triangle have lengths equal to the magnitudes of vector 𝐅, 𝐅 sub one, and 𝐅 sub two. If we label this triangle 𝐴𝐵𝐶 as shown, then using the properties of parallel lines and alternate angles, we know that angle 𝐴𝐵𝐶 is equal to 37 degrees. And since angles in a triangle sum to 180 degrees, then angle 𝐴𝐶𝐵 is equal to 27 degrees.

The law of sines, or sine rule, states that in any triangle 𝑎 over sin 𝐴 is equal to 𝑏 over sin 𝐵, which is equal to 𝑐 over sin 𝐶, where lowercase 𝑎, 𝑏, and 𝑐 are the lengths of the sides of our triangle and capital 𝐴, 𝐵, and 𝐶 are the measures of the angles opposite the corresponding sides. In our diagram, the force component 𝐅 sub two is opposite 116 degrees, 𝐅 sub one is opposite the angle of 37 degrees, and the force 𝐅 is opposite the angle 27 degrees. We are told in the question that the magnitude of this force 𝐅 is 99 newtons.

Substituting in these values gives us the equation 𝐅 sub two over sin of 116 degrees is equal to 𝐅 sub one over sin 37 degrees, which is equal to 99 over sin 27 degrees. In order to calculate the value of 𝐅 sub one, we will use the second and third parts of this equation. Multiplying through by the sin of 37 degrees gives us 𝐅 sub one is equal to 99 multiplied by sin of 37 degrees divided by sin of 27 degrees. Typing this into our calculator gives us 131.2355 and so on. We are asked to give our answer to two decimal places. Rounding up, we see that 𝐅 sub one is equal to 131.24 newtons.

We can repeat this process to find the magnitude of 𝐅 sub two. This time, we will use the first and third parts of our equation and multiply through by the sin of 116 degrees. 99 multiplied by sin of 116 degrees divided by sin of 27 degrees is equal to 195.9966 and so on. Rounding this to two decimal places gives us 196 or 196.00. The magnitude of 𝐅 sub two is 196.00 newtons to two decimal places.