- 7th Grade Work - Please help, I really need it T-T
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Answer:
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° <em>Substitute values in.</em>
- 138°+ ∠C =180° <em>Add</em>
- 138°-138° ∠C= 180°-138° <em>Subtract 138 from both sides.</em>
- ∠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° <em>Angles D and E are congruent</em>
- 42°+ 2∠D= 180° <em>Substitute 42 in for angle C</em>
- 42°-42° +2∠D= 180°-42° <em>Subtract 42 from both sides.</em>
- 2∠D= 138°
- 2∠D/2= 138°/2 <em>Divide both sides by 2.</em>
- ∠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° <em>Substitute 69 for angle CED</em>
- 69°-69° +∠DEB= 180°-69° <em>Subtract 69 from both sides</em>
- ∠DEB=111°
Angle DEB is equal to <u>111 degrees</u>