Question

1. given: if an angle measures more than 90°, then it is obtuse. given: m∠p = 115° conclusion: ∠p is obtuse valid by: law of detachment 24. given: if max gets an a on his geometry exam, then he will pass the class. given: if max does not get an a on his geometry exam, then he will repeat the class. conclusion: if max passes the class, then he will not repeat the class. valid by: law of syllogism

Explanation:

Problem 23

Step1: Identify the conditional statement

The conditional statement is "If an angle measures more than \(90^\circ\), then it is obtuse" (let \(p\): an angle measures more than \(90^\circ\), \(q\): it is obtuse, so \(p
ightarrow q\)).

Step2: Check the given fact

We are given \(m\angle P = 115^\circ\), so \(p\) is true for \(\angle P\) (since \(115^\circ>90^\circ\)).

Step3: Apply the Law of Detachment

The Law of Detachment states that if \(p
ightarrow q\) is true and \(p\) is true, then \(q\) is true. Here, \(p
ightarrow q\) is given, \(p\) ( \(m\angle P = 115^\circ>90^\circ\)) is true, so \(q\) ( \(\angle P\) is obtuse) is true. So the conclusion is valid by the Law of Detachment.

Step1: Define the statements

Let \(p\): Max gets an A on his Geometry exam, \(q\): Max will pass the class, \(r\): Max will repeat the class. The given statements are \(p
ightarrow q\) and \(
eg p
ightarrow r\). The conclusion is \(q
ightarrow
eg r\).

Step2: Analyze the logical flow

We know that either \(p\) or \(
eg p\) is true (Law of Excluded Middle). If \(p\) is true, then \(q\) is true (from \(p
ightarrow q\)) and \(r\) is false (since \(p\) is true, \(
eg p\) is false, so the second statement \(
eg p
ightarrow r\) doesn't apply in the sense of making \(r\) true; also, passing the class and repeating the class are mutually exclusive in this context). If \(
eg p\) is true, then \(r\) is true (from \(
eg p
ightarrow r\)) and \(q\) is false. So overall, if \(q\) is true (Max passes), then \(p\) must be true (since \(q\) is true only if \(p\) is true from \(p
ightarrow q\)), and if \(p\) is true, \(
eg p\) is false, so \(r\) is false (because \(
eg p
ightarrow r\) requires \(
eg p\) to be true for \(r\) to be true). So we can chain the implications: from \(p
ightarrow q\) and \(
eg p
ightarrow r\), we can infer \(q
ightarrow
eg r\) by considering the contrapositive and the mutual exclusivity, which is an application of the Law of Syllogism (in a more extended logical chain sense, or by recognizing the transitive and complementary nature of the implications).

Step3: Confirm the Law of Syllogism application

The Law of Syllogism allows us to chain implications. Here, we can think of the relationships between \(p\), \(q\), and \(r\) to form the conclusion \(q
ightarrow
eg r\) from the given \(p
ightarrow q\) and \(
eg p
ightarrow r\) by considering the logical consistency of the outcomes (passing vs. repeating). So the conclusion is valid by the Law of Syllogism (or a related logical syllogism principle for exclusive or - like situations).

Answer:

Law of Detachment

Problem 24