Question number 27. In the picture here, the segment PT is tangent to the circle C, which is centered at the point O, and the segment PS here bisects the angle TPR. Our job is to calculate the measure of angle TSP. So let me mark that angle in red. That's this angle here, TSP, and we know that from geometry the measure of the angle TSP is related to the arcs that this angle cuts away. So one arc is here, and the other arc opposite of that angle is here. So let's name this point here X, let's name this point here Y, and then we can say that the measure of angle TSP is equal to one-half of the sum of those arcs. So I will say XR here with a hat like that for that arc, plus TY with a hat for the second arc. And we need more relationships here between these, so we can look at the measure of angle P, so that's this angle here in blue, and it is also related to the measures of the angles of the arcs that it cuts away. So there is a far arc and a close one, so the measure of angle P, now that's equal to one-half, but we have the difference of those arcs from R to T minus from T to W, where W is this point over here. And let's do one more, we need here half the angle P, that's, we know that's half the angle P, the problem tells us that that segment P to S is an angle bisector, and the relationship here between the arcs would be this far arc and the close arc is from T to Y. So we have one-half minus, here we have X, T minus TY. And so we see right away that we have a relationship here between them, but let's work with the angle TSP first. So let me just multiply by two and say twice the measure of the angle TSP is equal to the sum of the arcs XR with TY, and then I will replace here, I will keep XR, and for TY I will use the relationship in the orange equation, so here the one-halves can be cancelled, and we can solve for TY, so TY would be the arc XT minus the measure of angle P. And then I'll use the equation in blue, so we have now XR plus XT minus one-half RT minus TW, like that. And then we can simplify a little bit, we have to notice first of all that the arc XR together with the arc XT, XR is here, XT is here, that's exactly the arc RT. So this is equal to now the arc RT minus half of that, minus half of RT, plus one-half of TW. So simplifying that we have one-half RT plus TW. And where is RT plus TW? So RT is here and then TW is here. So from R to W along these two arcs we have exactly a semicircle, half of that circle, so that arc has to measure exactly pi, so we have one-half here times pi, or pi over two. So we conclude that finally what we have is twice the measure of the angle TSP is equal to pi over two. So we can simply divide by two and the measure of that angle in red, TSP, is pi over four, or that is the same as 45 degrees. So we have our answer and that comes out to letter B, 45 degrees.

geometry

angle TSP

circle tangent

arc relationships

angle bisector