Question

Circle O with tangent MN. mSN= 60⁰; mQS=x; mQP=x+40; mPN=2x-16

Answer 1

Since MN is tangent to circle O, we know that the tangent and the radius at the point of tangency are perpendicular. Therefore, we can draw a radius from the center of the circle to the point of tangency:

[asy]
pair O, M, N;
O = (0,0);
M = (1,0);
N = rotate(60)*(1,0);
draw(Circle(O,1));
draw(M--N);
draw(O--N);
label("$O$",O,SE);
label("$M$",M,SE);
label("$N$",N,NW);
[/asy]

Let's call the center of the circle point O, and let's call the point where the tangent line MN intersects the radius ON point P:

[asy]
pair O, M, N, P;
O = (0,0);
M = (1,0);
N = rotate(60)*(1,0);
P = (1/2,sqrt(3)/2);
draw(Circle(O,1));
draw(M--N);
draw(O--N);
draw(O--P);
label("$O$",O,SE);
label("$M$",M,SE);
label("$N$",N,NW);
label("$P$",P,NE);
[/asy]

Since ON is a radius of the circle, we know that OP is also a radius and has length 1. Also, since triangle OMN is equilateral, we know that NP is half of a side, or (1/2) times the diameter of the circle, which means NP = 1/2.

Let's use these facts to find the measures of angles QSN and QSP. We know that QSP is a straight angle, so mQSP = 180. Also, since triangle QSN is isosceles, we know that mQSN = mQNS, which means mQSP = mQSN + mQNS. But mQSN is half of a side of an equilateral triangle, which means mQSN = 60. Therefore, we have:

mQSP = 180
mQSP = mQSN + mQNS
mQSN = 60

From the first two equations, we can solve for mQNS:

mQSP - mQSN = mQNS
180 - 60 = mQNS
mQNS = 120

Now let's use these angles to find x and then use x to find mPN. Since QNSP is a cyclic quadrilateral (because its vertices all lie on the circle), we know that opposite angles add up to 180. Therefore, we have:

mQNS + mQSP = 180
120 + 180 = mQSP
mQSP = 300

This tells us that:

mQS + mQSP + mPSN = 360
mQS + 300 + 60 = 360
mQS + 360 = 360
mQS = 0

Therefore, x = mQS = 0, and:

mPN = 2x-16
mPN = 2(0)-16
mPN = -16