Question

unit 3 part 1 assessment retake! round two – let’s do this! the retake is your chance to learn from your first attempt, show your growth, and level up your score. review your notes, go over the questions you missed, and come in ready to rock it. every step forward is progress, so give it your best shot and show what you’ve got! deltamath 11 - unit 3 part 1 | assessment retake 2024 credit until december 16 at 11:59 complete: 30% step 1 2 3 4 5 angle m∠dce = 32° m∠ade = 59° m∠dec = 90° m∠cde = 58° m∠adc = 117° (hover here to see on figure) m∠dcb = 117° reason given given perpendicular lines form right angles sum of angles in a triangle angle addition postulate corresponding angles in the diagram below, \\(\overline{ab} \parallel \overline{cd}\\), \\(\overline{ad} \parallel \overline{bc}\\), \\(\overline{de} \perp \overline{ce}\\), \\(m\angle dce = 32^\circ\\) and \\(m\angle ade = 59^\circ\\). find \\(m\angle ecb\\).

Explanation:

Step1: Analyze given angles

We know \( m\angle DCE = 32^\circ \), \( m\angle ADE = 59^\circ \), \( DE\perp CE \) (so \( m\angle DEC = 90^\circ \)), \( AD\parallel BC \), \( AB\parallel CD \). First, find \( m\angle CDE \): in \( \triangle DEC \), sum of angles is \( 180^\circ \), so \( m\angle CDE = 180^\circ - 90^\circ - 32^\circ = 58^\circ \) (sum of angles in a triangle).

Step2: Find \( m\angle ADC \)

\( m\angle ADC = m\angle ADE + m\angle CDE = 59^\circ + 58^\circ = 117^\circ \) (angle addition postulate).

Step3: Use parallel lines (corresponding angles)

Since \( AD\parallel BC \), \( m\angle DCB = m\angle ADC = 117^\circ \) (corresponding angles). Now, to find \( m\angle ECB \), we know \( m\angle DCB = m\angle DCE + m\angle ECB \), so \( m\angle ECB = m\angle DCB - m\angle DCE = 117^\circ - 32^\circ = 85^\circ \)? Wait, no, wait the problem might have a typo or maybe I misread. Wait, the step in the image shows \( m\angle DCB = 117^\circ \), and we need \( m\angle ECB \). Wait, \( \angle DCB = \angle DCE + \angle ECB \), so \( m\angle ECB = 117^\circ - 32^\circ = 85^\circ \)? But let's check again. Wait, maybe the correct approach: since \( AB\parallel CD \) and \( AD\parallel BC \), the figure is a parallelogram? Wait, no, let's re - express. Wait, the given steps in the image: step 5 has \( m\angle DCB = 117^\circ \), and we need \( m\angle ECB \). So \( m\angle ECB = m\angle DCB - m\angle DCE = 117^\circ - 32^\circ = 85^\circ \). Wait, but maybe I made a mistake. Wait, let's re - calculate:

Wait, \( DE\perp CE \), so \( \angle DEC = 90^\circ \). In \( \triangle DEC \), \( \angle DCE = 32^\circ \), so \( \angle CDE = 180 - 90 - 32 = 58^\circ \). Then \( \angle ADE = 59^\circ \), so \( \angle ADC = 59 + 58 = 117^\circ \). Since \( AD\parallel BC \), \( \angle ADC = \angle DCB = 117^\circ \) (corresponding angles). Then \( \angle ECB = \angle DCB - \angle DCE = 117 - 32 = 85^\circ \).

Wait, but maybe the problem is to find \( m\angle ECB \), and the steps lead to that. Let's confirm:

1. \( m\angle DCE = 32^\circ \) (given)
2. \( m\angle ADE = 59^\circ \) (given)
3. \( DE\perp CE \Rightarrow m\angle DEC = 90^\circ \) (perpendicular lines form right angles)
4. \( m\angle CDE = 180 - 90 - 32 = 58^\circ \) (sum of angles in a triangle)
5. \( m\angle ADC = 59 + 58 = 117^\circ \) (angle addition postulate)
6. \( AD\parallel BC \Rightarrow m\angle DCB = m\angle ADC = 117^\circ \) (corresponding angles)
7. \( m\angle ECB = m\angle DCB - m\angle DCE = 117 - 32 = 85^\circ \)

Wait, but the image shows a step where \( m\angle DCB = 117^\circ \), and we need to find \( m\angle ECB \). So the final answer for \( m\angle ECB \) is \( 85^\circ \)? But maybe I misread the problem. Wait, the original problem says "Find \( m\angle ECB \)". So following the steps:

Answer:

\( 85^\circ \) (assuming the correct calculation as above)