 In this video, we provide the solution to question number 19 from the practice final exam for math 1060 In which case given this diagram, which I should mention the circle in consideration is in fact the unit circle notice We have two examples of radii of this circle being length one So this is in fact the unit circle and so using this unit circle Diagram we need to prove that tangent of theta where theta is this angle in consideration here is equal to the length Ae so we're trying to show that this distance right here is the same thing as tangent of theta What are we allowed to use to prove that well first of all we know? This is the unit circle so this is a radius. This is a radius Those are length one and since we have a point on the unit circle We know that the x-coordinate here is cosine of theta and the y-coordinate is sine of theta as Illustrated on these two sides of a right triangle and so the basis of this argument can be around similar triangles So consider the triangle AOC right here. You're always going to use this triangle and what are we going to compare it to? Well, we need to use it right triangle that has Ae in it where you have Ae right here In which case that seems to suggest that we should use this triangle right here. All right That was kind of sloppy there. We're using this triangle right here And so we then claim and this is how the proof is going to start here. So note That the triangle AOC is similar to the triangle. Well, how does it compare right and we have to make sure the vertices connect with each other correctly, right? So notice they theta the angle coincides with both of these and there's also a right angle of a right here The right angle is there in there as well So when you do this correspondence angle a in the original triangle right here This is the complement of angle theta for which that would have to be angle E right here So we see that the vertex a coincides with E. What about oh oh coincides with theta Which then would be oh again, right? So you get oh like so and then the right angle which for the original triangles at C Coincided with a like I said earlier in which case we get the following the triangle AOC is similar to the triangle E. Oh a I put those in order to make the following statements a lot easier thus We get by proportions that a e over Right, so notice a is the side. We're trying to consider right here. What does a e coincide with a e on this one So with respect to I'm actually gonna draw these triangles to the side for a moment. Here's angle theta, right? So this is OCA with respect to angle theta. This is the opposite side This is the adjacent side and this is the hypotenuse Like so and then if we do this triangle right here again, I'm just gonna draw it the way it is on the screen They don't like so oh E and a so with respect to theta. This is the opposite side This is the adjacent side and this is the hypotenuse that also can help you figure out Which vertices coincide with each other so a e with respect to theta is the opposite side So that will coincide with a c which is the opposite side with respect to the other theta triangle All right, so we want to compare a e with a c but what else do we do well? What can go with a e? We have to pick another side of the triangle which oh e we don't really know anything about it It's longer than one I guess but oh a right here. It's a radius of the unit circle So that's the one I want to know so oh a right there now oh a Notice here is the adjacent side of the triangles. We have to compare that to the adjacent side over here For that triangle which would be OC like so so now using the diagram. What can we do here a e? We don't know oh a is a radius of the unit circle. So that is one. What about a c a C is equal to here sine of theta and Then OC is equal to cosine of theta Like though like so and so simplifying that we end up with a e is equal to sine theta over cosine theta Which is tangent theta and so therefore we've now proven the statement that a e is equal to tangent of theta