Question

find ( mangle krl ).
( mangle krl=square^{circ} )

Explanation:

Step1: Identify the total degrees around point R

The sum of angles around a point is \(360^{\circ}\). The right - angle (\(\angle RPM = 90^{\circ}\)), \(\angle HRQ=56^{\circ}\), and \(\angle KRP = 90^{\circ}\).

Step2: Use the angle - sum formula

We know that \(m\angle KRL+m\angle LRM + m\angle RPM+m\angle HRQ+m\angle KRH=360^{\circ}\). Since \(\angle KRH = 90^{\circ}\) and \(\angle LRM\) and \(\angle HRQ\) are vertical angles (so \(\angle LRM=\angle HRQ = 56^{\circ}\)), we can also use the fact that \(m\angle KRL+90^{\circ}+56^{\circ}+90^{\circ}=180^{\circ}\) (because \(\angle KRL\), \(\angle LRM\), \(\angle RPM\) and \(\angle HRQ\) are part of a straight - line and right - angle combinations. Another way:
The four angles \(\angle KRL\), \(\angle LRM\) (\( = 56^{\circ}\)), \(\angle RPM\) (\(=90^{\circ}\)) and \(\angle HRQ\) (\( = 56^{\circ}\)) and \(\angle KRH\) (\(=90^{\circ}\)) sum to \(360^{\circ}\). But a simpler approach is using the fact that \(\angle KRL\) is complementary to the angle equal to \(\angle HRQ\) (because of vertical angles and right - angles).
We know that \(m\angle KRL+90^{\circ}+56^{\circ}=180^{\circ}\) (the sum of angles in a half - circle ( \(180^{\circ}\)) with a right - angle (\(90^{\circ}\)) and an angle equal to \(\angle HRQ\) (vertical angles)).

$$m\angle KRL=180^{\circ}-90^{\circ}- 56^{\circ}$$

$$m\angle KRL = 34^{\circ}$$

Answer:

\(34\)