Question

day 4 - thursday, september 11th
if ∠x + ∠z = 90° and ∠y + ∠z = 90°, what can you conclude about ∠x and ∠y?
day 5 - friday, september 12th
quick proof sketch: if ∠p and ∠q are right angles, and ∠p and ∠r are supplementary, prove ∠q ≅ ∠r. (just outline steps, not full proof.)

Explanation:

Step1: Analyze first problem

Given $\angle X+\angle Z = 90^{\circ}$ and $\angle Y+\angle Z = 90^{\circ}$. We can express $\angle X=90^{\circ}-\angle Z$ and $\angle Y = 90^{\circ}-\angle Z$.

Step2: Conclude about angles

Since $\angle X=90^{\circ}-\angle Z$ and $\angle Y = 90^{\circ}-\angle Z$, we have $\angle X=\angle Y$.

Step3: Analyze second - problem

$\angle P$ and $\angle Q$ are right - angles, so $\angle P=\angle Q = 90^{\circ}$. Also, $\angle P$ and $\angle R$ are supplementary, so $\angle P+\angle R=180^{\circ}$.

Step4: Find measure of $\angle R$

Since $\angle P = 90^{\circ}$ and $\angle P+\angle R=180^{\circ}$, then $90^{\circ}+\angle R=180^{\circ}$, so $\angle R = 90^{\circ}$.

Step5: Prove angle congruence

Since $\angle Q = 90^{\circ}$ and $\angle R = 90^{\circ}$, we have $\angle Q\cong\angle R$.

Answer:

For the first problem: $\angle X=\angle Y$.
For the second problem: Outline of steps: 1. Note $\angle P=\angle Q = 90^{\circ}$ as they are right - angles. 2. Use the fact that $\angle P+\angle R=180^{\circ}$ (supplementary). 3. Solve for $\angle R$ to get $\angle R = 90^{\circ}$. 4. Conclude $\angle Q\cong\angle R$ since they have the same measure.