Question

in the rectangle below, $jn = 3x + 8$, $ln = 5x + 4$, and $m \angle nlm = 49^\circ$. find $kn$ and $m \angle nkl$.

Explanation:

Step1: Recall properties of rectangles

In a rectangle, the diagonals are equal in length and bisect each other. So, \( JN = LN = KN = MN \) (since diagonals bisect each other, \( JN = LN \) and \( KN = JN \) as diagonals are equal and bisect).
Given \( JN = 3x + 8 \) and \( LN = 5x + 4 \), set them equal:
\( 3x + 8 = 5x + 4 \)

Step2: Solve for \( x \)

Subtract \( 3x \) from both sides:
\( 8 = 2x + 4 \)
Subtract 4 from both sides:
\( 4 = 2x \)
Divide by 2:
\( x = 2 \)

Step3: Find \( KN \)

First, find \( JN \) (or \( LN \)) by substituting \( x = 2 \) into \( JN = 3x + 8 \):
\( JN = 3(2) + 8 = 6 + 8 = 14 \)
Since \( KN = JN \) (diagonals bisect each other in a rectangle), \( KN = 14 \)

Step4: Find \( m\angle NKL \)

In rectangle \( JKLM \), \( \angle NLM = 49^\circ \). Since \( KL \parallel JM \) and \( ML \) is a transversal, \( \angle KLM = 90^\circ \) (rectangle has right angles). Also, \( \triangle NKL \) is isosceles ( \( KN = LN = KL? \) Wait, no: \( KN = LN \), but \( KL \) is a side. Wait, \( \angle NLM = 49^\circ \), and \( \angle KLM = 90^\circ \), so \( \angle NKL \):
In \( \triangle NKL \), \( KN = LN \), but actually, \( \angle NLM = 49^\circ \), and \( \angle KNL \) is vertical to \( \angle JNM \), but better: \( \angle NKL = 90^\circ - 49^\circ = 41^\circ \)? Wait, no. Wait, \( \angle NLM = 49^\circ \), \( \angle KLM = 90^\circ \), so \( \angle KLN = 90^\circ - 49^\circ = 41^\circ \). But \( KN = LN \), so \( \triangle NKL \) is isosceles with \( KN = LN \)? Wait, no, \( KN = LN \) (diagonals bisect), so \( \triangle NKL \) has \( KN = LN \)? Wait, no, \( KN = JN \) and \( LN = JN \), so \( KN = LN \), so \( \triangle NKL \) is isosceles with \( KN = LN \), so \( \angle NKL = \angle NLK \). Wait, \( \angle NLK = 41^\circ \)? Wait, \( \angle NLM = 49^\circ \), \( \angle KLM = 90^\circ \), so \( \angle NLK = 90^\circ - 49^\circ = 41^\circ \). Thus, \( \angle NKL = 41^\circ \) (since \( KN = LN \), base angles are equal? Wait, no, \( KN = LN \), so \( \angle NKL = \angle NLK \). Wait, \( \angle NLK = 41^\circ \), so \( \angle NKL = 41^\circ \). Wait, let's check:
In rectangle, \( ML \perp KL \), so \( \angle KLM = 90^\circ \). \( \angle NLM = 49^\circ \), so \( \angle NLK = 90^\circ - 49^\circ = 41^\circ \). Since \( KN = LN \), \( \triangle NKL \) is isosceles with \( KN = LN \), so \( \angle NKL = \angle NLK = 41^\circ \).

Answer:

\( KN = \boldsymbol{14} \)
\( m\angle NKL = \boldsymbol{41}^\circ \)