7.3k views
1 vote
If AACB = ADCE, ZBAC = 3x-10,ZECD = 45°, and ZEDC = 2x+10.x= [?]

If AACB = ADCE, ZBAC = 3x-10,ZECD = 45°, and ZEDC = 2x+10.x= [?]-example-1
User Jun
by
6.4k points

1 Answer

1 vote

We are given that :


\begin{gathered} \Delta\text{ ACB }\cong\Delta\text{ DCE} \\ \cong\text{ means congruent} \end{gathered}
\begin{gathered} \text{The given angles :} \\ \angle\text{ BAC = 3x - 10} \\ \angle ECD=45^0 \\ \angle\text{ EDC = 2x + 10} \end{gathered}

Solution

The triangles ACB and DCE are congruent by AAA and SSS.

Hence,


\angle\text{ BAC = }\angle\text{ EDC }

Substituting and simplifying, we have:


\begin{gathered} 3x\text{ - 10 =2x + 10} \\ \text{Collect like terms} \\ x\text{ = 10 + 10} \\ x\text{ = 20} \end{gathered}

Answer: x = 20

User Michal Dobrodenka
by
7.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.