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In a rectangle MPKN, the diagonals intersect each other at point O. A line segment

OA

is an altitude of ΔMOP,

m∠AOP = 15°. Find m∠ONK.

Verified Answer :

taskmasters

It will really be helpful in your solution if you draw the rectangle, its diagonal and the altitude of ΔMOP. 

By doing so, you will find that ∠MOP is equal to twice m∠AOP and that is equal to 30°. Then, m∠MOP is a vertical angle of m∠NOK which means that they are equal. Therefore, m∠NOK is also 30°. 

We know that the sum of the angles of a triangle is equal to 180°.
                m∠NOK + m∠OKN + m∠ONK = 180°
And that m∠OKN = m∠ONK
so,
                 m∠NOK + 2m∠ONK = 180°
Substituting,
                 30° + 2m∠ONK = 180°
Hence,
                        m∠ONK = 75°

The sum of all the angle of a triangle is equal to the 180 degrees. The measure of the \(\angle ONK\) is 75 degrees.

Given information-

The rectangle MPKN shown in the attached image below.

The diagonals of the rectangle MPKN intersect each other at point.

The line segment OA is altitude of triangle MOP.

The measure of angle AOP is 15 degree.

Angle of triangle-

The sum of all the angle of a triangle is equal to the 180 degrees.

For the triangle the measure of \(\angle AOP\) is 15 degree given in the problem. As the line segment OA is the altitude \(\Delta MOP\) thus it the \(\angle OAP\) is right angle with measure 90 degrees.

As in triangle the sum of all the angles is equal to the 180 degrees. Thus,

\(\angle OPA+\angle OAP +\angle AOP=180\\\angle OPA+90+15=180\\\angle OPA=180-105\\\angle OPA=75\)

The line segment PM and KN are parallel for the rectangle. The PN makes the same angle at both ends. Thus,

\(\angle ONK=\angle AOP\\\angle ONK=75\)

Thus the measure of the \(\angle ONK\) is 75 degrees.

Learn more about the angle of triangle here;

https://brainly.com/question/10242960

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