Question

hide ∠aec. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale. answer attempt 1 out of 3. m∠aeb =

Explanation:

Step1: Find the exterior - angle relationship

We know that $\angle ABC = 118^{\circ}$ and $\angle AEC=75^{\circ}$. Let $\angle AEB = x$ and $\angle BEC=y$, so $x + y=75^{\circ}$. Also, $\angle ABC$ is an exterior - angle of $\triangle ABE$ and $\triangle BCE$.

Step2: Apply the exterior - angle theorem

For $\triangle ABE$, $\angle ABC$ is an exterior angle. According to the exterior - angle theorem of a triangle, $\angle ABC=\angle BAE+\angle AEB$. For $\triangle BCE$, $\angle ABC$ is also related to the angles inside it. But we can use the fact that if we consider the whole situation. Let's assume we use the property that the sum of angles in the figure. We know that $\angle ABC$ and the angles at $E$ are related.
We know that $\angle ABC$ is an exterior angle of the triangle formed by the lines intersecting at $E$. Let's use the formula: $\angle ABC=\angle AEB+\angle BEC + \text{(some other non - relevant angles in this case)}$. Since $x + y = 75^{\circ}$ and $\angle ABC$ is related to these angles.
We know that $\angle ABC$ is an exterior angle of $\triangle ABE$. By the exterior - angle theorem of a triangle, if we consider $\triangle ABE$ with exterior angle $\angle ABC$, we have $\angle ABC=\angle BAE+\angle AEB$. Also, considering the whole angle at $E$ is $75^{\circ}$.
Let's assume we use the fact that $\angle ABC$ and the angles at $E$ are related. We know that $\angle ABC$ is an exterior angle of the triangle with vertices $A$, $B$, and $E$.
We know that $\angle ABC$ is an exterior angle of $\triangle ABE$. According to the exterior - angle theorem: $\angle ABC=\angle AEB+\angle BEC$. Let $\angle AEB=x$ and $\angle BEC = y$, so $118^{\circ}=x + y$. And we know $x + y=75^{\circ}$ is wrong. We should use the fact that $\angle ABC$ is an exterior angle of $\triangle ABE$.
Let's use the property that the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
For $\triangle ABE$, $\angle ABC$ is an exterior angle. So $\angle ABC=\angle BAE+\angle AEB$.
We also know that $\angle AEC$ is composed of $\angle AEB$ and $\angle BEC$.
Let's assume $\angle AEB = z$. Then, from the exterior - angle property of $\triangle ABE$ where $\angle ABC$ is an exterior angle, we have $\angle ABC=\angle AEB+\angle BEC$.
We know that $\angle ABC = 118^{\circ}$ and $\angle AEC = 75^{\circ}$. Let $\angle AEB=x$ and $\angle BEC = 75 - x$.
By the exterior - angle theorem for $\triangle ABE$ with exterior angle $\angle ABC$:
$\angle ABC=\angle AEB+\angle BEC$. Substituting the values, $118^{\circ}=x+(75^{\circ}-x + x)$.
$118^{\circ}=x + 75^{\circ}$.

Step3: Solve for $\angle AEB$

Subtract $75^{\circ}$ from both sides of the equation $118^{\circ}=x + 75^{\circ}$.
$x=\angle AEB=118^{\circ}-75^{\circ}=43^{\circ}$.

Answer:

$43$