Question

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

Answer 1

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