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

Answer 1

Since tangent MN and line SP share a common point P, we know that angle MPS is a right angle.

Using the fact that the angles in a triangle sum to 180 degrees, we can find angle SPN as follows:

angle SPN = 180 - angle MPS - angle NPS
angle SPN = 180 - 90 - angle NPS
angle SPN = 90 - angle NPS

Next, we can use the fact that angles that intercept the same arc are congruent to each other:

angle QPN = angle NPS

Since angle QPN and angle SPN both intercept the arc PN, they must be congruent:

angle QPN = angle SPN
angle QPN = 90 - angle NPS

Now we have two equations that both equal angle QPN:

angle QPN = angle NPS
angle QPN = 90 - angle SPN

Setting these two equations equal to each other, we can solve for angle NPS:

angle NPS = 90 - angle SPN
angle NPS = 90 - (90 - angle QPN)
angle NPS = angle QPN

So angle NPS is equal to angle QPN, which we can find using the fact that angles inscribed in the same arc are congruent:

angle QPN = (1/2) m(QON)
angle QPN = (1/2) (m(QOS) - m(NOS))
angle QPN = (1/2) (m(QS) - m(NS))
angle QPN = (1/2) (x+40 - 60)
angle QPN = (1/2) (x-20)

Now we can set the two equations for angle QPN equal to each other and solve for x:

angle QPN = angle SPN
(1/2) (x-20) = 90 - (2x-16)
(x-20) = 180 - 4x + 32
3x = 128
x = 42.67

Now we can find angle NPS:

angle NPS = (1/2) (x-20)
angle NPS = (1/2) (42.67-20)
angle NPS = 11.34

Therefore, angle NPS is approximately 11.34 degrees.