Question

find ( mangle nql ) in parallelogram ( lmnq ).

( mangle nql = ) (\boxed{100}) ( ^circ ) (marked with x, incorrect attempt)

explain your reasoning.

by the parallelogram consecutive angles theorem (checked), ( angle nql ) and ( angle qnm ) (marked with x) are supplementary (checked).

Explanation:

Step1: Use parallelogram angle property

In parallelogram $LMNQ$, opposite angles are equal, so $m\angle QLM = m\angle QNM = 100^\circ$. Also, consecutive angles are supplementary, but we use triangle angle sum for $\triangle QOL$.

Step2: Identify known angles in $\triangle QOL$

We know $\angle OQL = 29^\circ$, and $\angle QLO$ is part of $\angle QLM$. Since $LM \parallel QN$, alternate interior angles give $\angle QLO = \angle LQN = 29^\circ$? No, correct: In $\triangle QOL$, $\angle OLQ$ is the angle at $L$ for the triangle, which is $\angle QLM$ minus the adjacent angle? No, better: In parallelogram, $LQ \parallel MN$, $LM \parallel QN$. The angle at $L$ is $100^\circ$, so $\angle QLP = 100^\circ$, and in $\triangle QPL$, we know $\angle PQL=29^\circ$, so $\angle QPL = 180-100-29=51^\circ$, but we need $\angle NQL$. Wait, no: $\angle NQL$ is the angle at $Q$ between $QN$ and $QL$. Since $LM \parallel QN$, $\angle QLM + \angle LQN = 180^\circ$ (consecutive interior angles).

Step3: Calculate $\angle NQL$

$\angle LQN = 180^\circ - m\angle QLM$
$\angle LQN = 180^\circ - 100^\circ = 80^\circ$
Wait, no, the $29^\circ$ is $\angle OQL$, so $\angle NQL = \angle LQN = 80^\circ$? No, correct: $\angle NQL$ is the angle we need. In $\triangle QON$, no, the correct approach: Consecutive angles in parallelogram are supplementary, so $\angle LQN$ (which is $\angle NQL$) and $\angle QLM$ are consecutive angles, so $m\angle NQL = 180^\circ - 100^\circ = 80^\circ$. The $29^\circ$ is a sub-angle in the triangle formed by diagonals, but the question uses the Consecutive Angles Theorem, so $\angle NQL$ and $\angle QLM$ are supplementary.

for reasoning blank:
By the Parallelogram Consecutive Angles Theorem, $\angle NQL$ and $\boldsymbol{\angle QLM}$ are supplementary. Since $m\angle QLM = 100^\circ$, $m\angle NQL = 180^\circ - 100^\circ = 80^\circ$.

Answer:

$m\angle NQL = 80^\circ$