We are given a circle with inscribed angle $\angle L = 62^\circ$. We are asked to find the measure of the inscribed angle $\angle N$. We know that $\overline{KN}$ is a diameter of the circle.

1. Problem Description

We are given a circle with inscribed angle $\angle L = 62^\circ$. We are asked to find the measure of the inscribed angle $\angle N$. We know that $\overline{KN}$ is a diameter of the circle.

2. Solution Steps

Since $\overline{KN}$ is a diameter, the inscribed angle $\angle L$ subtends a semicircle.

Since the sum of angles in a triangle is $180^\circ$, we can say that the angle $\angle LKN = 90^\circ$ because it subtends the semicircle.

Also, since $\overline{KN}$ is the diameter, we have $\angle LKN = 90^\circ$.

The sum of the angles in $\triangle KLN$ is $180^\circ$. Therefore,

$\angle L + \angle N + \angle K = 180^\circ$.

$62^\circ + \angle N + 90^\circ = 180^\circ$.

$\angle N = 180^\circ - 62^\circ - 90^\circ$.

$\angle N = 180^\circ - 152^\circ$.

$\angle N = 28^\circ$.

3. Final Answer

The measure of angle $N$ is $28^\circ$.