The sum of the interior angles of a triangle adds up to 180°.
Based on this, we can do the following...
0. Finding m∠C:
[tex]m\angle C+m\angle B+m\angle A=180[/tex][tex]m\angle C=180-m\angle B-m\angle A[/tex]
With the description of the problem, we know that m∠B = 48° and m∠A = 90°. Replacing these values:
[tex]m\angle C=180-48-90[/tex][tex]m\angle C=42[/tex]
With this angle and based on the same logic that the addition of the interior angles of a circle adds up 180°, we can get m∠CED. Also, as CD is congruent to CE, m∠CED = m∠CDE.
[tex]m\angle C+m\angle CED+m\angle CDE=180[/tex][tex]m\angle C+m\angle CED+m\angle CED=180[/tex][tex]m\angle C+2m\angle CED=180[/tex][tex]m\angle CED=\frac{(180-m\angle C)}{2}[/tex]
Replacing the value of m∠C previously calculated:
[tex]m\angle CED=\frac{(180-42)}{2}=\frac{138}{2}[/tex][tex]m\angle CED=69[/tex]
Finally, as we know segment CB is a straight line, the angle is 180°. Thus...
[tex]m\angle DEB+m\angle CED=180[/tex][tex]m\angle DEB=180-m\angle CED[/tex]
Replacing the value previously calculated:
[tex]m\angle DEB=180-69[/tex]
Answer:
[tex]m\angle DEB=111[/tex]
