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in the figure, $overrightarrow{ba}$ and $overrightarrow{bc}$ are opposi…

Question

in the figure, $overrightarrow{ba}$ and $overrightarrow{bc}$ are opposite rays. $overrightarrow{bh}$ bisects $angle ebc$. if $mangle ebc = 3r + 10$ and $mangle abe = 2r - 20$, find $mangle ebf$.

Explanation:

Step1: Use linear - pair property

Since $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are opposite rays, $\angle ABE+\angle EBC = 180^{\circ}$. So, $(2r - 20)+(3r + 10)=180$.

Step2: Simplify the equation

Combine like - terms: $2r+3r-20 + 10=180$, which gives $5r-10 = 180$.

Step3: Solve for $r$

Add 10 to both sides: $5r=180 + 10=190$. Then divide both sides by 5: $r=\frac{190}{5}=38$.

Step4: Find $m\angle EBC$

Substitute $r = 38$ into the expression for $m\angle EBC$: $m\angle EBC=3r + 10=3\times38+10=114 + 10=124^{\circ}$.

Step5: Use angle - bisector property

Since $\overrightarrow{BH}$ bisects $\angle EBC$, and we assume $\angle EBF=\frac{1}{2}\angle EBC$ (assuming $F$ is on the bisector $\overrightarrow{BH}$). So $m\angle EBF=\frac{1}{2}\times124^{\circ}=62^{\circ}$.

Answer:

$62^{\circ}$