We are given a circle with two intersecting chords, $\overline{PR}$ and $\overline{LQ}$. We are given that $m\angle L = 50^\circ$ and $m\angle R = 56^\circ$. We want to find $m\angle 1$, where $\angle 1$ is formed by the intersection of the two chords.

1. Problem Description

We are given a circle with two intersecting chords, $\overline{PR}$ and $\overline{LQ}$. We are given that $m\angle L = 50^\circ$ and $m\angle R = 56^\circ$. We want to find $m\angle 1$, where $\angle 1$ is formed by the intersection of the two chords.

2. Solution Steps

We can find the measure of angle 1 using the following formula:

$m\angle 1 = \frac{1}{2} (m\widehat{LR} + m\widehat{PQ})$

Since $m\angle L$ is an inscribed angle that intercepts arc $\widehat{QR}$, $m\widehat{QR} = 2 \cdot m\angle L = 2 \cdot 50^\circ = 100^\circ$.

Since $m\angle R$ is an inscribed angle that intercepts arc $\widehat{PQ}$, $m\widehat{PQ} = 2 \cdot m\angle R = 2 \cdot 56^\circ = 112^\circ$.

$m\angle 1 = \frac{1}{2} (m\widehat{LR} + m\widehat{PQ})$

Also $\angle 1$ and $\angle PLQ$ are vertical angles, so the arcs can be $m\widehat{LR}$ and $m\widehat{PQ}$

$m\angle LQR = m\angle R = 56^{\circ}$, because both are inscribed angles that intercept arc $\widehat{LQ}$.

$m\angle 1 = \frac{1}{2}(m\widehat{PQ} + m\widehat{LR}) = \frac{1}{2}(2 \cdot m\angle R + 2 \cdot m\angle L) = m\angle R + m\angle L$.

$m\angle 1 = 56^\circ + 50^\circ = 106^\circ$.

Another approach: Since $\angle 1$ is an exterior angle of $\triangle PRL$, its measure is equal to the sum of the measures of the two non-adjacent interior angles. Thus, $m\angle 1 = m\angle L + m\angle R = 50^\circ + 56^\circ = 106^\circ$.

3. Final Answer

106