Question

proof complete the paragraph proof of corollary 5.2.
a. case 1
given: $\triangle mno$; $angle m$ is a right angle.
prove: there can be at most one right angle in a triangle.
proof: it is given that in $\triangle mno$, $angle m$ is a right angle. $mangle m + mangle n + mangle o = 180^circ$ by the select choice $\boldsymbol{vee}$ $mangle m =$ select choice $\boldsymbol{vee}^circ$ by the definition of a select choice $\boldsymbol{vee}$ angle. so, select choice $\boldsymbol{vee}^circ + mangle n + mangle o = 180^circ$ by substitution and $mangle n + mangle o = 180^circ -$ select choice $\boldsymbol{vee}$ by the subtraction property of equality because $angle n$ and $angle o$ are angles in a triangle, $mangle n > 0^circ$ and $mangle o > 0^circ$. therefore, because $mangle o = 180^circ - mangle n -$ select choice $\boldsymbol{vee}$, by the substitution property of equality, $mangle n$ select choice $\boldsymbol{vee} 90^circ$ and $mangle o$ select choice $\boldsymbol{vee} 90^circ$. so, there cannot be two right angles in a triangle.

Explanation:

Step1: Recall Triangle Angle Sum

The sum of angles in a triangle is \(180^\circ\) (Triangle Angle Sum Theorem). So the first "Select Choice" (for the theorem) is "Triangle Angle Sum Theorem".

Step2: Right Angle Measure

A right angle (\(\angle M\)) has measure \(90^\circ\) (definition of a right angle). So \(m\angle M = 90^\circ\).

Step3: Substitute \(m\angle M\)

Substitute \(m\angle M = 90^\circ\) into \(m\angle M + m\angle N + m\angle O = 180^\circ\). We get \(90^\circ + m\angle N + m\angle O = 180^\circ\).

Step4: Solve for \(m\angle N + m\angle O\)

Subtract \(90^\circ\) from both sides: \(m\angle N + m\angle O = 180^\circ - 90^\circ = 90^\circ\) (Subtraction Property of Equality).

Step5: Analyze Angles \(\angle N\) and \(\angle O\)

Since \(\angle N\) and \(\angle O\) are angles in a triangle, they must be positive (greater than \(0^\circ\)). So neither can be \(90^\circ\) (because their sum is \(90^\circ\), and if one were \(90^\circ\), the other would be \(0^\circ\), which isn't a valid angle). Thus, only \(\angle M\) is a right angle.

Answer:

1. First "Select Choice" (theorem): Triangle Angle Sum Theorem
2. \(m\angle M\): \(90\)
3. After substitution: \(90 + m\angle N + m\angle O = 180\) (then simplify to \(m\angle N + m\angle O = 90\))
4. Conclusion: Only \(\angle M\) is a right angle, so a triangle has at most one right angle.

(Note: Filling in the "Select Choice" boxes:

• First box: Triangle Angle Sum Theorem
• \(m\angle M\) box: \(90\)
• Then, \(m\angle N + m\angle O = 90^\circ\), so neither \(\angle N\) nor \(\angle O\) can be \(90^\circ\) (as their sum is \(90^\circ\) and they are positive), proving at most one right angle.)