# Question Video: Finding the Missing Term in the Vector Forces Expression Mathematics

Complete the following: In the given figure, 𝐅₁ = (𝐑 × …)/(sin (𝜃₁ + 𝜃₂)).

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### Video Transcript

Complete the following. In the given figure, 𝐅 sub one is equal to 𝐑 multiplied by what divided by the sin of 𝜃 sub one plus 𝜃 sub two.

The diagram shows a parallelogram of forces, where the sides of the parallelogram are equal to the magnitudes of vectors 𝐅 sub one and 𝐅 sub two and the diagonal of the parallelogram is the resultant of these forces 𝐑. In order to find an expression for 𝐅 sub one, we can begin by finding the missing angles in triangle 𝑂𝐴𝐶. Angles 𝐵𝑂𝐶 and 𝑂𝐶𝐴 are alternate angles, meaning that angle 𝑂𝐶𝐴 is equal to 𝜃 sub two.

We know that angles in a triangle sum to 180 degrees. This means that the measure of angle 𝑂𝐴𝐶 is equal to 180 degrees minus 𝜃 sub one plus 𝜃 sub two. We can now use the sine rule or law of sines to find an expression for 𝐅 sub one in terms of 𝐑 together with 𝜃 sub one and 𝜃 sub two. The law of sines states that 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 three sides of our triangle and capital 𝐴, 𝐵, and 𝐶 are the measures of the angles opposite the corresponding sides.

In triangle 𝑂𝐴𝐶, side length 𝑂𝐴 is opposite the angle at 𝐶. This gives us 𝐅 sub one over sin of 𝜃 sub two. The resultant 𝐑 is opposite the angle equal to 180 degrees minus 𝜃 sub one plus 𝜃 sub two. Substituting this into the law of sines gives us 𝐑 over sin of 180 degrees minus 𝜃 sub one plus 𝜃 sub two. Recalling the symmetry of the sine function, we know that the sin of 180 degrees minus 𝛼 is equal to sin 𝛼. This means that the right-hand side of our equation can be rewritten as 𝐑 over sin of 𝜃 sub one plus 𝜃 sub two. In order to make 𝐅 sub one the subject of our equation, we can multiply through by sin of 𝜃 sub two. 𝐅 sub one is therefore equal to 𝐑 multiplied by sin of 𝜃 sub two divided by sin of 𝜃 sub one plus 𝜃 sub two.

And we can therefore conclude that the missing part of the equation in the question is sin of 𝜃 sub two.