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Question
1 - 2 congruent segments and angles homework
- if qr bisects ∠mqr, qn ⊥ lm and m∠mqk = 34°, and m∠pqn = 22°, find each measure.
m∠nqm =
m∠kqr =
m∠mqr =
m∠lqr =
m∠pql =
m∠lqn =
Step1: Recall angle - bisector property
If a ray bisects an angle, it divides the angle into two equal angles. Given that $QK$ bisects $\angle MQR$ and $m\angle MQK = 34^{\circ}$, then $m\angle KQR=m\angle MQK = 34^{\circ}$. So $m\angle MQR=m\angle MQK + m\angle KQR=34^{\circ}+34^{\circ}=68^{\circ}$.
Step2: Use perpendicular - angle property
Since $QN\perp LM$, $\angle NQM$ and $\angle MQR$ are complementary (because $\angle NQL = 90^{\circ}$). So $m\angle NQM=90^{\circ}-m\angle MQR$. Substituting $m\angle MQR = 68^{\circ}$, we get $m\angle NQM=90^{\circ}-68^{\circ}=22^{\circ}$.
Step3: Use angle - bisector property for $\angle PQN$
Given that $m\angle PQN = 22^{\circ}$, and $\angle PQL$ and $\angle PQN$ are complementary (because $\angle NQL = 90^{\circ}$), so $m\angle PQL=90^{\circ}-m\angle PQN=90^{\circ}-22^{\circ}=68^{\circ}$.
Step4: Find $\angle LQR$
$\angle LQR=\angle PQL+\angle PQN+\angle NQM+\angle MQR$. Since $\angle PQL = 68^{\circ}$, $\angle PQN = 22^{\circ}$, $\angle NQM = 22^{\circ}$, $\angle MQR = 68^{\circ}$, then $m\angle LQR=68^{\circ}+22^{\circ}+22^{\circ}+68^{\circ}=180^{\circ}$.
Step5: Find $\angle LQN$
Since $\angle NQL = 90^{\circ}$, $m\angle LQN = 90^{\circ}$.
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$m\angle MQR = 68^{\circ}$
$m\angle KQR = 34^{\circ}$
$m\angle NQM = 22^{\circ}$
$m\angle LQR = 180^{\circ}$
$m\angle PQL = 68^{\circ}$
$m\angle LQN = 90^{\circ}$