Answer:
In the given diagram, QP is tangent to a circle with centre O, RS is a straight line, T is a point on the circle, and PS bisects TPQ. We know that SPQ = 22°. Let's try to find out the value of the angle TPQ.
Since QP is tangent to the circle, the angle between RS and QP (i.e., angle RQP) is equal to the angle between QP and the radius drawn to the point of tangency (i.e., angle QOT). So, we can say that:
angle RQP = angle QOT
Also, since PS bisects TPQ, we can say that:
angle TPS = angle TPQ / 2
Now, let's consider the triangle TPQ. We know that:
angle TPQ + angle TQP + angle PTQ = 180° [Sum of angles in a triangle]
Substituting the values we have:
angle TPQ + angle TQP + (angle TPS + angle SPQ) = 180°
angle TPQ + angle TQP + (angle TPQ/2 + 22°) = 180°
Multiplying both sides by 2 to eliminate the fraction:
2(angle TPQ) + 2(angle TQP) + angle TPQ + 44° = 360°
Simplifying:
3(angle TPQ) + 2(angle TQP) = 316°
We don't know the values of angle TPQ and angle TQP, so we can't solve this equation exactly. However, we do know that these angles are both less than 180° (since they are angles in a triangle). Therefore, we can try some values for angle TPQ (let's call it x) and see if we can find a corresponding value for angle TQP that satisfies the equation.
If we take x = 40°, then we get:
3(40°) + 2(angle TQP) = 316°
120° + 2(angle TQP) = 316°
2(angle TQP) = 196°
angle TQP = 98°
Now, we can use the fact that angle TPS = angle TPQ / 2 to find angle TPS:
angle TPS = x/2 = 20°
Finally, we can use the fact that PS bisects TPQ to find angle PQT:
angle PQT = angle TPS
Explanation: