Question

given that $\angle a \cong \angle b$, gavin conjectured that $\angle a$ and $\angle b$ are complementary angles. which statement is a counterexample to gavin’s conjecture? \bigcirc $m\angle a = 30^\circ$ and $m\angle b = 60^\circ$ \bigcirc $m\angle a = 10^\circ$ and $m\angle b = 15^\circ$ \bigcirc $m\angle a = 25^\circ$ and $m\angle b = 25^\circ$ \bigcirc $m\angle a = 45^\circ$ and $m\angle b = 45^\circ$

Explanation:

A counterexample to a conjecture is an example that shows the conjecture is false. Gavin's conjecture is that if \( \angle A \cong \angle B \) (so \( m\angle A = m\angle B \)), then \( \angle A \) and \( \angle B \) are complementary (their measures add up to \( 90^\circ \)). We need to find a case where \( m\angle A = m\angle B \) but \( m\angle A + m\angle B
eq 90^\circ \).

• For the first option: \( 30^\circ

eq 60^\circ \), so \( \angle A
ot\cong \angle B \), not a counterexample.

• For the second option: \( 10^\circ

eq 15^\circ \), so \( \angle A
ot\cong \angle B \), not a counterexample.

• For the third option: \( m\angle A = m\angle B = 25^\circ \), so \( \angle A \cong \angle B \). Their sum is \( 25^\circ + 25^\circ = 50^\circ

eq 90^\circ \), so this shows the conjecture is false.

• For the fourth option: \( m\angle A = m\angle B = 45^\circ \), sum is \( 90^\circ \), which supports the conjecture, not a counterexample.

Answer:

\( m\angle A = 25^\circ \) and \( m\angle B = 25^\circ \)