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

Answer 1

Since MN is tangent to circle O, we know that mSN is a right angle (a radius and tangent are always perpendicular). Therefore, mNSO is a right triangle, and we can use trigonometry to find the value of x:

tan(60) = SN/NS
sqrt(3) = SN/NS
SN = sqrt(3)NS

Now consider triangle QPN. We know that mPN = 2x-16, and we can find mPQN by using the fact that the angles of a triangle sum to 180:

mPQN = 180 - mNQP - mNPQ
mNQP = mQS = x
mNPQ = mQPN = mSNQ = 90 - mSNP = 90 - 60 = 30

mPQN = 180 - x - 30 = 150 - x

Next, we can use the Law of Sines to find the length of PN:

PN/sin(mPQN) = QP/sin(mQPN)
PN/sin(150-x) = QP/sin(30)
QP = PN*sin(30)/(sin(150-x))

We also know that QP = QS + SP, and since QS is a radius of circle O and SP is tangent to circle O at N, we have:

QP = OS - OP = r - OP

So we can combine these equations to get:

r - OP = PN*sin(30)/(sin(150-x))

OP = r - PN*sin(30)/(sin(150-x))

Finally, we can use the Law of Cosines to find the measure of arc PN:

cos(mPNQ) = (PN^2 + QP^2 - PQ^2)/(2*PN*QP)

mPNQ = 2*cos^-1((PN^2 + QP^2 - PQ^2)/(2*PN*QP))

Substituting in our expressions for QP and OP, we get:

mPNQ = 2*cos^-1((PN^2 + (r - PN*sin(30)/(sin(150-x)))^2 - r^2)/(2*PN*(r - PN*sin(30)/(sin(150-x)))))

This is a messy expression, but we can simplify by expanding and collecting like terms:

mPNQ = 2*cos^-1((PN^2 - 2*PN*r*sin(30)/(sin(150-x)) + r^2 - PN^2*sin^2(30)/(sin^2(150-x)) - r^2)/(2*PN*(r - PN*sin(30)/(sin(150-x)))))

Canceling some terms, we get:

mPNQ = 2*cos^-1((r - PN*sin(30)/(sin(150-x)) - PN*cos(30)/(sin(150-x)))/(r - PN*sin(30)/(sin(150-x))))

Now we can substitute in our values and simplify:

mPNQ = 2*cos^-1((r - (2x-16)*sin(30)/(sin(150-x)) - (2x-16)*cos(30)/(sin(150-x)))/(r - (2x-16)*sin(30)/(sin(150-x))))
mPNQ = 2*cos^-1((2r - (2x-16)*sqrt(3) - (2x-16))/2r)

Canceling a factor of (2x-16), we get:

mPNQ = 2*cos^-1((r - 8*sqrt(3) + 8)/r)
mPNQ = 2*cos^-1((r + 8 - 8*sqrt(3))/r)

So the measure of arc PN is twice this angle:

arc PN = 2*mPNQ = 2*acos((r + 8 - 8*sqrt(3))/r)