Question

note: figure not drawn to scale.
in the figure, $ac = cd$. the measure of angle $ebc$ is $45^{circ}$, and the measure of angle $acd$ is $104^{circ}$. what is the value of $x$?

Explanation:

Step1: Find angles in isosceles triangle ACD

Since $AC = CD$, in $\triangle ACD$, let $\angle CAD=\angle CDA = y$. Using the angle - sum property of a triangle ($\angle ACD+\angle CAD+\angle CDA = 180^{\circ}$), and $\angle ACD = 104^{\circ}$, we have $104^{\circ}+y + y=180^{\circ}$, which simplifies to $2y=180^{\circ}- 104^{\circ}=76^{\circ}$, so $y = 38^{\circ}$.

Step2: Use exterior - angle property

$\angle BCE$ is the supplement of $\angle ACD$, so $\angle BCE=180^{\circ}-\angle ACD=180 - 104=76^{\circ}$. In $\triangle BCE$, we know that $\angle EBC = 45^{\circ}$ and $\angle BCE = 76^{\circ}$. By the exterior - angle property of a triangle, the exterior angle $x$ of $\triangle BCE$ is equal to the sum of the two non - adjacent interior angles. So $x=\angle EBC+\angle BCE$.

Step3: Calculate the value of x

$x = 45^{\circ}+76^{\circ}=121^{\circ}$

Answer:

$121$