Question

If AACBZECD=AADCE, ZCAB = 63⁰,52°, and ZDEC = 5xDEx = [?]CB

Answer 1

given:

$\angle CAB=63^0,\text{ }\angle ECD=52^0\text{ , }\angle DEC=5x$

Find: x

Explanation:

$\begin{gathered} \angle ECD=\angle BCA=52^0 \\ \angle CAB+\angle ABC+\angle BCA=180^0 \\ 63^0+\angle ABC+52^0=180^0 \\ \angle ABC=65^0 \end{gathered}$$\begin{gathered} \angle ABC=\angle CDE \\ \angle CDE+\angle DEC+\angle ECD=180^0 \\ 65^0+5x+52^0=180^0 \\ 5x=63^0 \\ x=12.6^0 \end{gathered}$