Question

In the rectangle MPKN, the diagonals intersect each other at point O. Segment OA is an altitude of triangle MOP, m<AOP=15 degrees. Find the measure of angle ONK.

Answer 1

In rectangle MPKN, the diagonals intersect each other at point O. We know that in a rectangle, the diagonals bisect each other and are equal in length. Therefore, point O is the midpoint of both diagonals.

Let's analyze triangle MOP where we have the following:

- \( \angle AOP = 15^\circ \), which is given as the angle between the altitude OA and the diagonal OP.
- Since OA is an altitude, \( \angle OAP = 90^\circ \).

To find the other angles, we can use the triangle angle sum property.

In triangle OAP, we have:
\[
\angle OAP + \angle AOP + \angle OPA = 180^\circ.
\]
Substituting the known values:
\[
90^\circ + 15^\circ + \angle OPA = 180^\circ,
\]
which simplifies to:
\[
\angle OPA = 180^\circ - 105^\circ = 75^\circ.
\]

Now we need to find the measure of angle ONK. We note that in rectangle MPKN, both diagonal pairs intersect at 90 degrees since the sides of a rectangle meet at 90 degrees.

Specifically, we have:
- \( \angle MOP = 90^\circ \) (the diagonals of the rectangle).
- Thus, each angle around point O can be represented as follows:

- Since \( \angle MOP + \angle NOP = 180^\circ \) and \( \angle MOP = 90^\circ \),
\[
\angle NOP = 90^\circ.
\]

Now, we focus on triangle NOK. Here, since the diagonals bisect the angles in a rectangle, we can say that:

- \( \angle NOK = 90^\circ \) (the same reason as above).

Since we are interested in \( \angle ONK\), we have \( \angle ONK + \angle NOK = 90^\circ \).

Now we place this together:
\[
\angle ONK = 90^\circ - \angle OAN.
\]

With \( \angle OAN = \angle OAP + \angle AON = 15 + 75 = 90 - 15 = 15^\circ \).

Thus:
\[
\angle ONK = 90^\circ - 15^\circ = 75^\circ.
\]

So, the measure of angle ONK is:
\[
\boxed{75^\circ}.
\]