Question

in the diagram below, $overline{be}congoverline{ec}$, $mangle aeb = 21^{circ}$ and $mangle ecd=118^{circ}$. find $mangle a$. you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find $\angle BEC$

$\angle AEB+\angle BEC = 180^{\circ}$ (linear - pair of angles). Given $\angle AEB = 21^{\circ}$, so $\angle BEC=180 - 21=159^{\circ}$.

Step2: Find $\angle EBC$ and $\angle ECB$ in $\triangle BEC$

Since $BE = EC$, $\triangle BEC$ is isosceles. Let $\angle EBC=\angle ECB = x$. Using the angle - sum property of a triangle ($\angle BEC+\angle EBC+\angle ECB = 180^{\circ}$), we have $159 + 2x=180$. Solving for $x$: $2x=180 - 159 = 21$, so $x = 10.5^{\circ}$.

Step3: Find $\angle ABC$

$\angle ABC = 180-\angle EBC$. So $\angle ABC=180 - 10.5 = 169.5^{\circ}$.

Step4: Find $\angle ACB$

$\angle ACB=\angle ECB = 10.5^{\circ}$.

Step5: Find $\angle A$ in $\triangle ABC$

Using the angle - sum property of a triangle ($\angle A+\angle ABC+\angle ACB = 180^{\circ}$). Substitute $\angle ABC = 169.5^{\circ}$ and $\angle ACB = 10.5^{\circ}$: $\angle A=180-(169.5 + 10.5)=0^{\circ}$ which is incorrect. Let's correct the third step. $\angle ABC=\angle EBC = 10.5^{\circ}$.
Using the angle - sum property of a triangle in $\triangle ABC$ ($\angle A+\angle ABC+\angle ACB = 180^{\circ}$), and $\angle ACB = 10.5^{\circ}$, $\angle ABC = 10.5^{\circ}$.
So $\angle A=180-(10.5 + 10.5)=159^{\circ}$.

Answer:

$159^{\circ}$