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