# Question Video: Finding the Magnitudes of Two Perpendicular Forces Mathematics

A force 𝐅, acting in a northerly direction, is the resultant of two forces 𝐅₁ and 𝐅₂. The force 𝐅₁ has a magnitude of 172 N and is acting 60° north of east, and the force 𝐅₂ acts in a westerly direction. Find the magnitudes of 𝐅 and 𝐅₂.

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### Video Transcript

A force 𝐅, acting in a northerly direction, is the resultant of two forces 𝐅 sub one and 𝐅 sub two. The force 𝐅 sub one has a magnitude of 172 newtons and is acting 60 degrees north of east, and the force 𝐅 sub two acts in a westerly direction. Find the magnitudes of 𝐅 and 𝐅 sub two.

We can see from the diagram that the forces 𝐅 and 𝐅 sub two are perpendicular to one another. This means that the angle between them is equal to 90 degrees. We can therefore create a right triangle of forces, where the lengths of each side of the triangle are equal to the magnitudes of the forces. Acting vertically upwards, we have the force 𝐅. Acting horizontally to the left, we have the magnitude of the 𝐅 sub two force. And the hypotenuse of our right triangle has length equal to the magnitude of 𝐅 sub one. We are told that this is equal to 172 newtons.

And since the force 𝐅 sub one acts at an angle 60 degrees north of east, we can add this angle to our right triangle. We can now use the sine and cosine trigonometric ratios to calculate the missing forces. These state that in any right triangle, the sin of angle 𝜃 is equal to the opposite over the hypotenuse and the cos of angle 𝜃 is the adjacent over the hypotenuse. As already mentioned, the hypotenuse is the side opposite the right angle. The side opposite the 60-degree angle has length equal to the magnitude of force 𝐅, and the side between the 60-degree angle and the right angle is the adjacent.

Substituting our values into the sin ratio gives us the sin of 60 degrees is equal to the magnitude of 𝐅 divided by 172. We can multiply both sides of the equation by 172. And since the sin of 60 degrees is root three over two, we have the magnitude of 𝐅 is equal to 86 root three newtons. Repeating this process with the cos ratio gives us the cos of 60 degrees is equal to the magnitude of 𝐅 sub two over 172. Once again, we can multiply through by 172, and the cos of 60 degrees is one-half. The magnitude of 𝐅 sub two is therefore equal to 86 newtons.

The magnitudes of 𝐅 and 𝐅 sub two are 86 root three newtons and 86 newtons, respectively.