what is m∠cbd?

m∠cbd = \boxed{}°

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Question

what is m∠cbd?

m∠cbd = \boxed{}°

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Accepted Answer

# Explanation:
## Step1: Identify congruent triangles
Since $ DC = DB = 44 $, and $ \angle C $ and $ \angle A $ are right angles, triangles $ DCB $ and $ DAB $ are congruent (HL theorem). Thus, $ \angle CBD=\angle ABD $.
## Step2: Set up angle equation
The angles at $ B $ satisfy $ 2s + (s + 24^\circ)= 90^\circ $ (since $ \angle CBA $ is a right angle? Wait, no, actually, since $ \triangle DCB \cong \triangle DAB $, $ \angle CBD = \angle ABD $, and $ \angle CBD + \angle ABD + $ wait, no, looking at the diagram, $ \angle CBD = 2s $ and $ \angle ABD = s + 24^\circ $, and since $ DC \perp BC $ and $ DA \perp BA $, and $ DC = DA = 44 $, so $ DB $ is the angle bisector? Wait, no, actually, triangles $ DCB $ and $ DAB $ are congruent (right triangles with hypotenuse $ DB $ and leg $ DC = DA $), so $ \angle CBD=\angle ABD $. Wait, no, the angles at $ B $: $ \angle CBD = 2s $, $ \angle ABD = s + 24^\circ $, and since $ \triangle DCB \cong \triangle DAB $, $ \angle CBD=\angle ABD $? Wait, no, maybe $ \angle CBD + \angle ABD = 90^\circ $? Wait, no, the diagram shows $ \angle C $ and $ \angle A $ are right angles, so $ BC \perp DC $ and $ BA \perp DA $, and $ DC = DA = 44 $, so $ DB $ is the angle bisector of $ \angle CBA $, so $ \angle CBD=\angle ABD $. Wait, that would mean $ 2s = s + 24^\circ $, so $ s = 24^\circ $, then $ \angle CBD = 2s = 48^\circ $. Wait, let's check: if $ s = 24 $, then $ \angle ABD = 24 + 24 = 48^\circ $, so $ \angle CBD = \angle ABD = 48^\circ $, and $ \angle CBA = 48 + 48 = 96^\circ $? No, that can't be. Wait, maybe $ \angle CBA $ is a right angle? No, the diagram has $ \angle C $ and $ \angle A $ as right angles, so $ BC $ and $ BA $ are not necessarily perpendicular. Wait, maybe the sum of $ \angle CBD $ and $ \angle ABD $ is equal to the angle between $ BC $ and $ BA $, but since $ DC = DA $ and $ DB $ is common, triangles $ DCB $ and $ DAB $ are congruent (HL), so $ BC = BA $, so $ \triangle CBA $ is isoceles? No, maybe the angles at $ B $ in the two right triangles: $ \angle CBD = 2s $, $ \angle ABD = s + 24^\circ $, and since the triangles are congruent, $ \angle CBD = \angle ABD $, so $ 2s = s + 24 $, so $ s = 24 $, then $ \angle CBD = 2*24 = 48^\circ $. Wait, that makes sense. So solve $ 2s = s + 24 $, so $ s = 24 $, then $ \angle CBD = 2s = 48^\circ $.
## Step3: Calculate $ s $
Set $ 2s = s + 24 $ (since $ \triangle DCB \cong \triangle DAB $, so $ \angle CBD = \angle ABD $). Subtract $ s $ from both sides: $ s = 24 $.
## Step4: Find $ m\angle CBD $
$ m\angle CBD = 2s = 2*24 = 48^\circ $.

# Answer:
$ 48 $

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QUESTION IMAGE

Question

Explanation:

Step1: Identify congruent triangles

Step2: Set up angle equation

Step3: Calculate \( s \)

Step4: Find \( m\angle CBD \)

Answer: