### Video Transcript

A force of magnitude 41 newtons
acts due south. It is resolved into two components
as shown on the diagram. Find the magnitudes of π
sub one
and π
sub two. Give your answer to two decimal
places.

Letβs begin by adding our force of
41 newtons to the diagram. It acts due south, so itβs this one
here. That force is itself then resolved
into two parts or components. Those are π
sub one and π
sub
two. Now, you might notice that these
two components are given in vector form. But actually, weβre trying to find
the magnitudes of these. The magnitude of the vector is
essentially its length. And so we want to find these two
dimensions here. And what we might try to do is add
right-angled triangles to our diagram.

But itβs not particularly easy to
do so here since the forces π
sub one and π
sub two are not acting at right angles
to one another. So instead, we draw something
called the parallelogram of forces. Weβre drawing lines parallel to π
sub one and π
sub two, therefore creating a parallelogram. The force of 41 newtons acting due
south then makes up the diagonal of this parallelogram. And so we can then use angle facts
to complete some of our missing measurements.

We know that alternate angles are
equal, so we can add a 45-degree angle here and a 60-degree angle here. We also know that angles in a
triangle add up to 180 degrees. So by subtracting 45 and 60 from
180, we find the missing two angles in our triangles. Letβs enlarge the triangle on the
left-hand side, linking the force of 41 newtons and π
sub two. We have a non-right-angled triangle
for which we know the measure of all three angles and one of the sides. This means that to find the measure
of a second side, we can use the law of sines.

Labeling our triangle as shown, we
see weβre going to link π over sin π΄ with π over sin π΅. Substituting what we know about our
triangle into the formula, and we see that the magnitude of π
sub two divided by
sin of 60 is equal to 41 over sin of 75. To solve this equation to find the
magnitude of π
sub two β notice Iβve written that with bars β we multiply by sin of
60. And so the magnitude of π
sub two
is 41 over sin of 75 times sin of 60, and thatβs equal to 36.759 and so on. Correct two decimal places then,
the magnitude of π
sub two is 36.76 newtons.

Weβre going to repeat this process
to find the magnitude of π
sub one. Now, to do this, we could repeat
the process of redrawing the triangle on the right-hand side. However, we know that this side,
the side labeled π΄π΅, on the triangle we drew is parallel to the side that
represents the vector component π
sub one. In fact, since this is a
parallelogram, itβs also of equal length. And so if we calculate the length
of line segment π΄π΅, that will tell us the magnitude of π
sub one. This time, weβre linking π and sin
π΅ with π and sin πΆ. And so our equation is 41 over sin
75 equals the magnitude of π
sub one over sin of 45.

To solve for π
sub one, weβre
going to multiply both sides by sin of 45 degrees. And so the magnitude of π
sub one
is 41 over sin of 75 times sin of 45, which is 30.014 and so on. Correct to two decimal places,
thatβs 30.01. And so the magnitude of π
sub one
is 30.01 newtons and the magnitude of π
sub two is 36.76 newtons.