### Video Transcript

If the angle between two tangents
drawn from an external point 𝑃 to a circle of radius 𝑎 and centre 𝑂 is 60
degrees, find the length of 𝑂𝑃.

This is a geometry problem. And so our first step is to draw a
diagram which represents the given information in our question. We want two tangents from an
external point 𝑃 to a circle.

Let’s first draw the circle. Here it is. Now let’s mark the external point
𝑃. That just means it’s a point
outside the circle, and you can mark it wherever you like. I’m going to mark it here. Now we need two tangents drawn from
this point 𝑃 to the circle. That means two lines which touch
the circle but don’t cross the circle and which pass through 𝑃. Here’s one of those lines and
here’s the other.

We’re told a bit more about the
circle. It has a radius of 𝑎, and its
centre is the point 𝑂. So we mark these facts in. And finally, we’re told that the
angle between these two tangents is 60 degrees. That’s this angle here. That’s all the given information in
the question.

What we are required to find is the
length of 𝑂𝑃. We can mark in the line segment
𝑂𝑃 on the diagram. We can call it unknown length
𝑥. Now we read the question one more
time to make sure we got all the information from it. If the angle between two tangents
drawn from an external point 𝑃 to a circle of radius 𝑎 and centre 𝑂 is 60
degrees, find the length of 𝑂𝑃.

Now we’ve got everything from the
question on the diagram, we can start to answer the question. Whenever we have a question
involving tangents to a circle, this theorem should come to mind. The tangent at any point of a
circle is perpendicular to the radius through the point of contact. So we can take the top tangent and
draw the radius from the point of contact. This theorem tells us that this
radius makes a right angle with the tangent. And we do exactly the same with the
other tangent. There’s a right angle here
also.

Now we can see that we have two
right-angled triangles which share the hypotenuse 𝑂𝑃. If we let the point of contact be
𝐴 and 𝐵, then these two triangles are 𝐴𝑂𝑃 and 𝐵𝑂𝑃. Let’s consider these triangles.

As mentioned before, they’re both
right-angled triangles; they have a right angle. They also share the hypotenuse 𝑂𝑃
whose length we have to find. And they have two other equal
sides. 𝑂𝐴 and 𝑂𝐵 are both radii of the
circle, with length 𝑎. These triangles both have a right
angle, they have the same hypotenuse, and they have equal sides. And so, by the RHS rule, they are
congruent.

In particular, this means that
corresponding angles and sides of these triangles are equal. The corresponding angles 𝐴𝑃𝑂 and
𝐵𝑃𝑂 must be equal. And we can see from the diagram
that their sum is 60 degrees, so each of these angles must be half 60 degrees, that
is, 30 degrees.

Now we can see by considering
either triangle 𝐴𝑂𝑃 or 𝐵𝑂𝑃 that we can find the length 𝑥 using
trigonometry. We have an angle of 30 degrees in
the right-angled triangle 𝐴𝑂𝑃. We know that the side opposite this
angle has a length of 𝑎. And we’re looking for the length of
the hypotenuse 𝑥.

In the triangle 𝐴𝑂𝑃, we see that
sin of 30 degrees is the opposite 𝑎 over the hypotenuse 𝑥. Let’s clear some room to solve this
for 𝑥. 30 degrees is a special angle, and
we know that sin of 30 degrees is a half. So we have that a half is 𝑎 over
𝑥. If we multiply both sides by 𝑥, we
get that 𝑥 over two is equal to 𝑎. And finally, multiplying by two on
both sides of the equation, we get that 𝑥 is two 𝑎. Remember that 𝑥 was the length of
𝑂𝑃 that we had to find. We found that this length of 𝑂𝑃
is two 𝑎. That’s twice the radius of the
circle.

The main steps we used to answer
this question were to represent all the information from the question in the
diagram. As the question involved tangents
to circles, we used the fact that the tangent at any point of a circle is
perpendicular to the radius through the point of contact. This allowed us to see two
right-angled triangles, which we then proved were congruent using the RHS rule. Using the fact that they were
congruent allowed us to find another angle in each triangle, which allowed us to use
trigonometry to find the unknown side length.