Question

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what is m∠acb?

m∠acb = \boxed{}°
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Explanation:

Step1: Identify Congruent Triangles

Triangles \(ABC\) and \(ADC\) are right triangles with \(AB = AD = 56\) and \(AC\) common. So, \(\triangle ABC \cong \triangle ADC\) (HL congruence). Thus, \(\angle ACB=\angle ACD\)? Wait, no, the angles at \(C\) are \(s + 24^\circ\) and \(4s\)? Wait, actually, since the triangles are congruent, the angles \(\angle ACB\) and \(\angle ACD\) should be equal? Wait, no, looking at the diagram, \(CB\) and \(CD\) are the legs, \(AB = AD = 56\), right angles at \(B\) and \(D\). So, \(AC\) is the hypotenuse, so \(\triangle ABC \cong \triangle ADC\) (HL: hypotenuse \(AC\), leg \(AB = AD\)). Therefore, the angles \(\angle ACB\) and \(\angle ACD\) are equal? Wait, no, the angles at \(C\) are given as \(s + 24^\circ\) (for \(\angle ACB\)) and \(4s\) (for \(\angle ACD\))? Wait, no, maybe I misread. Wait, actually, since the triangles are congruent, the angles \(\angle ACB\) and \(\angle ACD\) should be equal? Wait, no, the problem is that \(CB\) and \(CD\) are the other legs? Wait, no, \(AB\) and \(AD\) are both 56, right angles at \(B\) and \(D\), so \(CB = CD\) (since congruent triangles have equal legs). Wait, maybe the angles \(s + 24^\circ\) and \(4s\) are equal? Wait, that must be it. So, set \(s + 24 = 4s\).

Step2: Solve for \(s\)

Set \(s + 24 = 4s\). Subtract \(s\) from both sides: \(24 = 3s\). Then, \(s = 8\).

Step3: Find \(m\angle ACB\)

\(m\angle ACB = s + 24^\circ\). Substitute \(s = 8\): \(8 + 24 = 32^\circ\)? Wait, no, wait, \(4s\) would be \(32^\circ\) too? Wait, \(4s = 4\times8 = 32\), and \(s + 24 = 8 + 24 = 32\). Wait, that works. Wait, but wait, maybe I mixed up. Wait, the angle \(\angle ACB\) is \(s + 24\), and \(\angle ACD\) is \(4s\). Since the triangles are congruent, these angles should be equal. So, \(s + 24 = 4s\), so \(3s = 24\), \(s = 8\). Then, \(m\angle ACB = s + 24 = 8 + 24 = 32\)? Wait, no, wait, maybe \(\angle ACB\) is \(4s\)? Wait, no, the diagram shows \(\angle ACB\) as \(s + 24^\circ\) and \(\angle ACD\) as \(4s\). But since the triangles are congruent, \(\angle ACB = \angle ACD\), so \(s + 24 = 4s\), so \(s = 8\), then \(\angle ACB = 4s = 32\)? Wait, no, \(s + 24 = 32\), \(4s = 32\). So either way, the angle is \(32^\circ\)? Wait, no, wait, maybe I made a mistake. Wait, let's re-express.

Wait, the key is that \(\triangle ABC \cong \triangle ADC\), so corresponding angles \(\angle ACB\) and \(\angle ACD\) are equal. Therefore, \(s + 24 = 4s\). Solving: \(3s = 24\) → \(s = 8\). Then, \(m\angle ACB = s + 24 = 8 + 24 = 32\) degrees? Wait, no, \(4s = 32\), so if \(\angle ACB\) is \(4s\), then it's 32. Wait, maybe the label was wrong. Wait, the problem says "What is \(m\angle ACB\)?", and in the diagram, \(\angle ACB\) is labeled \(s + 24^\circ\), and \(\angle ACD\) is \(4s\). But since the triangles are congruent, those angles must be equal. So, \(s + 24 = 4s\), so \(s = 8\), then \(m\angle ACB = s + 24 = 32^\circ\). Wait, but let's check again.

Wait, maybe I messed up the congruence. Let's see: \(AB = AD = 56\), right angles at \(B\) and \(D\), \(AC\) is common. So by HL, \(\triangle ABC \cong \triangle ADC\). Therefore, \(CB = CD\) (corresponding sides), and \(\angle ACB = \angle ACD\) (corresponding angles). Therefore, \(\angle ACB = \angle ACD\), so \(s + 24 = 4s\). Solving for \(s\): \(3s = 24\) → \(s = 8\). Then, \(m\angle ACB = s + 24 = 8 + 24 = 32\) degrees. Wait, but \(4s = 32\) as well, so that's consistent. So the measure of \(\angle ACB\) is \(32^\circ\)? Wait, no, wait, maybe \(\angle ACB\) is \(4s\)? Wait, the diagram shows \(\angle ACB\) as \(s + 24^\circ\) and \(\angle ACD\) as \(4s\). But since the triangles are congruent, those angles are equal, so \(s + 24 = 4s\), so \(s = 8\), then \(\angle ACB = s + 24 = 32\), and \(\angle ACD = 4s = 32\). So that's correct.

Answer:

\(32\)