We are given a circle with center O and chord PT. We are given that $m(\angle PBT) = 70^\circ$. We need to find: a) $m(\angle TPO)$ b) $m(\text{arc } TQ)$

1. Problem Description

We are given a circle with center O and chord PT. We are given that $m(\angle PBT) = 70^\circ$. We need to find:

a) $m(\angle TPO)$

b) $m(\text{arc } TQ)$

2. Solution Steps

a) To find $m(\angle TPO)$, we need to find the relationship between $\angle PBT$ and $\angle POT$. Since $\angle PBT$ is the angle subtended by the chord PT at point B on the circumference, the angle subtended by the same chord at the center is twice the angle subtended at the circumference.

So, $m(\angle POT) = 2 \cdot m(\angle PBT) = 2 \cdot 70^\circ = 140^\circ$.

Now consider the triangle $\triangle POT$. Since OP and OT are both radii of the circle, $OP = OT$. Therefore, $\triangle POT$ is an isosceles triangle. This implies that $\angle OPT = \angle OTP$.

The sum of angles in a triangle is $180^\circ$. Therefore, in $\triangle POT$, we have

$m(\angle POT) + m(\angle OPT) + m(\angle OTP) = 180^\circ$.

Since $\angle OPT = \angle OTP$, let's call their measure $x$. So,

$140^\circ + x + x = 180^\circ$.

$2x = 180^\circ - 140^\circ = 40^\circ$.

$x = \frac{40^\circ}{2} = 20^\circ$.

So, $m(\angle OPT) = m(\angle OTP) = 20^\circ$.

Since $\angle TPO = \angle OPT$, we have $m(\angle TPO) = 20^\circ$.

b) To find $m(\text{arc } TQ)$, we know that $\angle TBQ$ is inscribed angle and $m(\angle PBT) = 70^\circ$.

$\angle TBP$ and $\angle TBQ$ form a line. Since $\angle PBT = 70^\circ$, then $\angle TBQ = 180^\circ - 70^\circ = 110^\circ$.

The angle at the center is twice the inscribed angle, so $m(\angle TOQ) = 2 \cdot m(\angle TBQ) = 2(110^\circ) = 220^\circ$.

The arc $TQ$ corresponds to the central angle $\angle TOQ$. So, $m(\text{arc } TQ) = m(\angle TOQ) = 220^\circ$.

Another way to consider the solution for b:

We know that $\angle TPO = 20^\circ$. The arc TP corresponds to the central angle $\angle TOP = 140^\circ$. The total degrees of circle $= 360^\circ$.

Then, $TQ = 360 - arc TP = 360 - 140 = 220^\circ$.

3. Final Answer

a) $m(\angle TPO) = 20^\circ$

b) $m(\text{arc } TQ) = 220^\circ$