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

Complete the following: In the given figure, πβ = (π Γ β¦)/(sin (πβ + πβ)).

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### Video Transcript

Complete the following. In the given figure, π sub one is equal to π multiplied by what divided by the sin of π sub one plus π sub two.

The diagram shows a parallelogram of forces, where the sides of the parallelogram are equal to the magnitudes of vectors π sub one and π sub two and the diagonal of the parallelogram is the resultant of these forces π. In order to find an expression for π sub one, we can begin by finding the missing angles in triangle ππ΄πΆ. Angles π΅ππΆ and ππΆπ΄ are alternate angles, meaning that angle ππΆπ΄ is equal to π sub two.

We know that angles in a triangle sum to 180 degrees. This means that the measure of angle ππ΄πΆ is equal to 180 degrees minus π sub one plus π sub two. We can now use the sine rule or law of sines to find an expression for π sub one in terms of π together with π sub one and π sub two. The law of sines states that π over sin π΄ is equal to π over sin π΅, which is equal to π over sin πΆ, where lowercase π, π, and π are the lengths of the three sides of our triangle and capital π΄, π΅, and πΆ are the measures of the angles opposite the corresponding sides.

In triangle ππ΄πΆ, side length ππ΄ is opposite the angle at πΆ. This gives us π sub one over sin of π sub two. The resultant π is opposite the angle equal to 180 degrees minus π sub one plus π sub two. Substituting this into the law of sines gives us π over sin of 180 degrees minus π sub one plus π sub two. Recalling the symmetry of the sine function, we know that the sin of 180 degrees minus πΌ is equal to sin πΌ. This means that the right-hand side of our equation can be rewritten as π over sin of π sub one plus π sub two. In order to make π sub one the subject of our equation, we can multiply through by sin of π sub two. π sub one is therefore equal to π multiplied by sin of π sub two divided by sin of π sub one plus π sub two.

And we can therefore conclude that the missing part of the equation in the question is sin of π sub two.