We need to find the measure of angle $R$, denoted as $m\angle R$. We are given that the segments that appear to be tangent are tangent. We are also given that the intercepted arc $PQ$ is $40^\circ$ and the angle formed by the two tangents outside the circle at point $V$ is $120^\circ$.

1. Problem Description

We need to find the measure of angle $R$, denoted as $m\angle R$. We are given that the segments that appear to be tangent are tangent. We are also given that the intercepted arc $PQ$ is $40^\circ$ and the angle formed by the two tangents outside the circle at point $V$ is $120^\circ$.

2. Solution Steps

Let $m\angle V = 120^\circ$ and $m\stackrel{\frown}{PQ} = 40^\circ$.

We have the formula for the angle formed by two tangents drawn to a circle from an exterior point:

$m\angle V = \frac{1}{2} (m\stackrel{\frown}{PTQ} - m\stackrel{\frown}{PQ})$

where $PTQ$ is the major arc and $PQ$ is the minor arc. The entire circle is $360^\circ$, so $m\stackrel{\frown}{PTQ} = 360^\circ - m\stackrel{\frown}{PQ} = 360^\circ - 40^\circ = 320^\circ$.

Then

$120^\circ = \frac{1}{2} (320^\circ - 40^\circ) = \frac{1}{2} (280^\circ) = 140^\circ$.

This is a contradiction! Therefore, we must consider the tangents to intersect at R.

Let $x$ be the $m\stackrel{\frown}{TQ}$. Then $m\angle R = \frac{1}{2}(m\stackrel{\frown}{PQ} - m\stackrel{\frown}{TQ})$

$40^\circ = \frac{1}{2} (m\stackrel{\frown}{PQ})$

so $m\stackrel{\frown}{PQ} = 2 * 40^\circ = 80^\circ$.

Also $m\stackrel{\frown}{PT} + m\stackrel{\frown}{TQ} + m\stackrel{\frown}{PQ} = 360^\circ$.

We also know that

$m\angle V = \frac{1}{2} (m\stackrel{\frown}{PT} - m\stackrel{\frown}{TQ})$

$120^\circ = \frac{1}{2} (m\stackrel{\frown}{PT} - m\stackrel{\frown}{TQ})$.

Thus, $240^\circ = m\stackrel{\frown}{PT} - m\stackrel{\frown}{TQ}$.

We have $m\stackrel{\frown}{PT} + m\stackrel{\frown}{TQ} + 40^\circ = 360^\circ$.

$m\stackrel{\frown}{PT} + m\stackrel{\frown}{TQ} = 320^\circ$.

So, $m\stackrel{\frown}{PT} = 320^\circ - m\stackrel{\frown}{TQ}$.

$240^\circ = 320^\circ - m\stackrel{\frown}{TQ} - m\stackrel{\frown}{TQ}$.

$2m\stackrel{\frown}{TQ} = 320^\circ - 240^\circ = 80^\circ$.

$m\stackrel{\frown}{TQ} = 40^\circ$.

Therefore $m\stackrel{\frown}{PT} = 320^\circ - 40^\circ = 280^\circ$.

The question asks for angle R.

$m\angle R = \frac{1}{2} (m\stackrel{\frown}{PT} - m\stackrel{\frown}{PQ}) = \frac{1}{2} (280^\circ - 80^\circ) = \frac{1}{2} (200^\circ) = 100^\circ$.

3. Final Answer

100