Question

in the diagram below, (overline{be}congoverline{ec}), (mangle bec = 50^{circ}) and (mangle a=42^{circ}). find (mangle aeb). some lines that appear straight are straight, but the figure is not otherwise drawn to scale. answer attempt 1 out of 2

Explanation:

Step1: Find base - angles of isosceles triangle $\triangle BEC$

Since $\overline{BE}\cong\overline{EC}$, in $\triangle BEC$, let $\angle EBC=\angle ECB = x$. Using the angle - sum property of a triangle ($\angle BEC + \angle EBC+\angle ECB=180^{\circ}$), we have $50^{\circ}+x + x=180^{\circ}$. So, $2x=180^{\circ}- 50^{\circ}=130^{\circ}$, and $x = 65^{\circ}$.

Step2: Find $\angle ABE$

$\angle ABE = 180^{\circ}-\angle EBC$. So, $\angle ABE=180^{\circ}-65^{\circ}=115^{\circ}$.

Step3: Find $\angle AEB$ in $\triangle ABE$

In $\triangle ABE$, we know $\angle A = 42^{\circ}$ and $\angle ABE = 115^{\circ}$. Using the angle - sum property of a triangle ($\angle A+\angle ABE+\angle AEB = 180^{\circ}$), we get $\angle AEB=180^{\circ}-\angle A-\angle ABE$. Substituting the values, $\angle AEB=180^{\circ}-42^{\circ}-115^{\circ}=23^{\circ}$.

Answer:

$23^{\circ}$