Question

9.
o is the center of the circle. assume that lines that appear to be tangent are tangent. what is the value of x?
(there is a diagram with a circle centered at o, a tangent line from p to q on the circle, and a line from p through the circle to o, forming a 29° angle at p and an angle x° at o between oq and op.)

Explanation:

Step1: Recall tangent - radius property

A tangent to a circle is perpendicular to the radius at the point of tangency. So, \(\angle OQP = 90^{\circ}\).

Step2: Use triangle angle - sum property

In triangle \(OQP\), we know that the sum of angles in a triangle is \(180^{\circ}\). We are given \(\angle OPQ=29^{\circ}\) and \(\angle OQP = 90^{\circ}\). Let \(\angle POQ=x\). Then, by the angle - sum property of a triangle (\(\angle OPQ+\angle OQP+\angle POQ = 180^{\circ}\)), we substitute the known values:
\(29^{\circ}+90^{\circ}+x = 180^{\circ}\)
First, calculate \(29^{\circ}+90^{\circ}=119^{\circ}\). Then the equation becomes \(119^{\circ}+x = 180^{\circ}\).
To find \(x\), we subtract \(119^{\circ}\) from both sides: \(x=180^{\circ}- 119^{\circ}=61^{\circ}\)? Wait, no, wait. Wait, actually, the angle between the two tangents from an external point to a circle: the line from the external point to the center bisects the angle between the tangents, but here we have one tangent (PQ) and one secant? Wait, no, PQ is tangent, and PO is a line from external point P to center O, intersecting the circle at some point. Wait, the correct property: the radius is perpendicular to the tangent, so \(\angle OQP = 90^{\circ}\), and in triangle OQP, we have angles: \(\angle P = 29^{\circ}\), \(\angle OQP=90^{\circ}\), so \(\angle POQ=180 - 90 - 29=61^{\circ}\)? Wait, no, that's wrong. Wait, actually, the angle between the tangent and the radius is \(90^{\circ}\), so triangle OQP is right - angled at Q. So \(\angle QOP=x\), \(\angle QPO = 29^{\circ}\), \(\angle OQP = 90^{\circ}\). Then \(x=90 - 29=61\)? No, wait, in a right - triangle, the two non - right angles sum to \(90^{\circ}\). Wait, \(\angle QPO+\angle QOP = 90^{\circ}\), because \(\angle OQP = 90^{\circ}\). So \(x = 90^{\circ}-29^{\circ}=61^{\circ}\)? Wait, no, that's not right. Wait, no, the angle between the tangent and the line from the external point to the center: the formula is that if a tangent from an external point P touches the circle at Q, and O is the center, then \(\angle OQP = 90^{\circ}\), and in triangle OPQ, \(\angle OPQ = 29^{\circ}\), so \(\angle POQ=90^{\circ}-29^{\circ}=61^{\circ}\)? Wait, no, sum of angles in a triangle is \(180\), so \(29 + 90+x=180\), so \(x = 180-(29 + 90)=180 - 119 = 61\). Wait, but actually, the correct approach: the angle between the tangent and the radius is \(90^{\circ}\), so \(\angle OQP = 90^{\circ}\). Then in \(\triangle OQP\), \(\angle P = 29^{\circ}\), \(\angle OQP = 90^{\circ}\), so \(\angle QOP=180-(29 + 90)=61^{\circ}\). Wait, but that seems low. Wait, no, maybe I mixed up the angle. Wait, the external angle: the angle between the tangent and the secant is equal to half the difference of the intercepted arcs. But here, PQ is tangent, PO is a line from P to O, so PQ is tangent at Q, so OQ is radius, OQ \(\perp\) PQ. So \(\triangle OQP\) is right - angled at Q. So angles: \(\angle P = 29^{\circ}\), \(\angle Q = 90^{\circ}\), so \(\angle O=180 - 90 - 29 = 61^{\circ}\). Wait, but that's what the calculation says. Wait, maybe I made a mistake. Wait, no, let's re - check. Sum of angles in a triangle: \(A + B + C=180\). Here, \(A = 29\), \(B = 90\), so \(C=180 - 29 - 90=61\). So \(x = 61\). Wait, but the problem is, is PO a secant? No, PQ is tangent, and PO is a line from P to O, passing through the circle (so it's a secant, intersecting the circle at two points: one is the point where PO meets the circle (let's say R) and O is the center. But the tangent at Q: OQ is radius, so OQ \(\perp\) PQ. So triangle OQP is right - angled at Q. So the angle at O is \(x = 180 - 90 - 29=61\). So the value of \(x\) is \(61\)? Wait, no, wait, I think I messed up. Wait, the correct formula: the angle between the tangent and the radius is \(90^{\circ}\), so \(\angle OQP = 90^{\circ}\), and the angle between the two tangents from P: but here there is only one tangent (PQ) and one secant (PO). Wait, no, PQ is tangent, and the other line from P to the circle is a secant. But the angle between tangent PQ and secant PO is equal to half the measure of the intercepted arc. The intercepted arc is arc Q (the minor arc). Wait, the angle between tangent PQ and secant PO is \(\angle QPO = 29^{\circ}\), and this angle is equal to half the measure of the intercepted arc (arc Q). Wait, no, the formula is: the measure of an angle formed by a tangent and a secant drawn from a point outside the circle is equal to half the difference of the measures of the intercepted arcs. But if we have a tangent and a secant, the angle is half the measure of the intercepted arc (the major arc minus the minor arc). Wait, no, when the secant passes through the center, then the intercepted arc: the tangent is PQ, secant is PO (passing through O, the center). So the intercepted arc is arc Q (the arc from Q to the other intersection point, but since PO passes through O, the other intersection point is the diametrically opposite? No, PO is a line from P to O, so it intersects the circle at one point (let's say R) and O is the center. So the two intercepted arcs are arc QR (the minor arc) and arc Q (the major arc). Wait, this is getting confusing. Let's go back to the triangle. Since PQ is tangent to the circle at Q, OQ is perpendicular to PQ, so \(\angle OQP = 90^{\circ}\). In triangle OQP, we have three angles: \(\angle QPO = 29^{\circ}\), \(\angle OQP = 90^{\circ}\), and \(\angle POQ=x\). By the angle - sum property of a triangle: \(\angle QPO+\angle OQP+\angle POQ = 180^{\circ}\). So \(29^{\circ}+90^{\circ}+x = 180^{\circ}\). So \(x=180^{\circ}-(29^{\circ}+90^{\circ})=180 - 119 = 61^{\circ}\). So the value of \(x\) is \(61\). Wait, but I think I made a mistake earlier in the formula. The correct way is using the right - triangle angle sum. So yes, \(x = 61^{\circ}\).

Answer:

\(61^{\circ}\)