Question

I WILL MARK BRAINLIEST PLEASE HELP FIND ANGLE DEB ITS DUE AT 12 PLEASE HELP PLEASE

Answer 1

Answer:

$\boxed {\boxed {\sf \angle DEB = 111 \textdegree}}$

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:

• ∠A+∠B+∠C= 180°

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