Video Transcript
In the given figure, find the
values of π₯ and π¦.
Letβs have a look at the diagram
weβve been given. There is a circle with center π
and then two tangents to this circle: the lines π΄π΅ and π΄πΆ. These two tangents intersect at a
point outside the circle, point π΄. And weβre told that the measure of
the angle formed between these two tangents is π₯ degrees.
Weβre also given expressions for
the measures of the two arcs intercepted by these tangents. The arc we can assume to be the
minor arc, π΅πΆ, has a measure of two π₯ degrees. And the measure of the major arc
π΅πΆ is π¦ degrees. In order to find the values of
these unknowns π₯ and π¦, we need to recall the relationship that exists between the
angle between two tangents and the measures of their intercepted arcs.
We know that the measure of the
angle between two tangents that intersect outside a circle is half the positive
difference of the measures of the intercepted arcs. We can therefore form an
equation. The measure of the angle between
the two tangents is π₯ degrees. The major intercepted arc is π¦
degrees, and the minor intercepted arc is two π₯ degrees. So we have the equation π₯ degrees
is equal to a half π¦ degrees minus two π₯ degrees.
Now as everything in this equation
is measured in degrees, we can remove the units. And we have π₯ is equal to a half
multiplied by π¦ minus two π₯. Now we canβt solve this equation
because we have two unknowns and only one equation, but we can manipulate it a
little.
Weβll begin by multiplying both
sides by two to give two π₯ is equal to π¦ minus two π₯. And then we can add two π₯ to each
side to give four π₯ is equal to π¦. Now as weβve already said, we canβt
solve this equation because we have two unknowns and only a single equation. But we have at least defined
explicitly what the relationship is between π¦ and π₯. In order to find the values of π₯
and π¦ though, we need a second equation.
Well, we also know that the measure
of a full circle is 360 degrees. Adding together the measures of the
major and minor arcs then, we can form a second equation, π¦ plus two π₯ is equal to
360. We now have two linear equations in
two unknowns. And so we can solve them
simultaneously. Our first equation π¦ equals four
π₯ gives an expression for π¦ in terms of π₯. So we can substitute this
expression for π¦ into our second equation, giving four π₯ plus two π₯ equals
360.
We can then group the like terms on
the left-hand side of the equation to give six π₯ equals 360 and finally divide both
sides of the equation by six to give π₯ equals 60. Weβve found the value then of one
of the two unknowns. In order to find the value of the
other, we need to substitute this value of π₯ into the other equation. That gives π¦ equals four
multiplied by 60, which is 240. A quick check confirms that this
value of π¦ plus twice this value of π₯ is indeed equal to 360.
So by recalling the angles of
intersecting tangents theorem, weβve solved the problem and found the values of the
two unknowns. π₯ is equal to 60, and π¦ is equal
to 240.