Answer:
[tex]\boxed {\boxed {\sf \angle DEB = 111 \textdegree}}[/tex]
Step-by-step explanation:
We want to find angle DEB. We know AB is perpendicular to AC, CD is congruent to CE and angle B is 48 degrees.
ABC is a triangle. The angles in a triangle must add up to 180 degrees. Therefore:
We know two angles: ∠B= 48° and ∠A= 90° (the little square denotes a right angle).
- 90°+48°+ ∠C= 180° Substitute values in.
- 138°+ ∠C =180° Add
- 138°-138° ∠C= 180°-138° Subtract 138 from both sides.
- ∠C= 42°
Note that angle C is part of another triangle. It is isosceles because the two legs (CD and CE) are congruent. Therefore, the two base angles (E and D) are congruent.
- ∠C+∠D+∠E= 180°
- ∠C+ 2∠D= 180° Angles D and E are congruent
- 42°+ 2∠D= 180° Substitute 42 in for angle C
- 42°-42° +2∠D= 180°-42° Subtract 42 from both sides.
- 2∠D= 138°
- 2∠D/2= 138°/2 Divide both sides by 2.
- ∠D= 69°
∠D and ∠E equal 69 degrees.
Angle CED (∠E) and DEB are on a straight line together. Therefore, they are supplementary and equal 180 degrees.
- ∠CED+ ∠DEB= 180
- 69° +∠DEB= 180° Substitute 69 for angle CED
- 69°-69° +∠DEB= 180°-69° Subtract 69 from both sides
- ∠DEB=111°
Angle DEB is equal to 111 degrees
