Question

(ma 912 gr 1.1)
in the figure, $overrightarrow{ca}$ and $overrightarrow{ce}$ are opposite rays and $overrightarrow{gc}$ bisects $angle bgd$
if $mangle bgc=(6x - 13)^{circ}$ and $mangle cgf=(4x + 3)^{circ}$, what is $mangle bgf?$
mangle bgf =

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{GC}$ bisects $\angle BGD$, then $m\angle BGC=m\angle CGD=(6x - 13)^{\circ}$.

Step2: Set up an equation using the angle - addition postulate

We know that $\angle BGF=\angle BGC+\angle CGF$. Also, $\angle CGD=\angle CGF$ (vertically opposite angles or by the fact that $\overrightarrow{GC}$ bisects $\angle BGD$ and the geometric relationship in the figure). So we set up the equation $6x - 13=4x + 3$.

Step3: Solve the equation for $x$

Subtract $4x$ from both sides: $6x-4x - 13=4x-4x + 3$, which simplifies to $2x-13 = 3$. Then add 13 to both sides: $2x-13 + 13=3 + 13$, so $2x=16$. Divide both sides by 2: $x = 8$.

Step4: Find $m\angle BGC$ and $m\angle CGF$

Substitute $x = 8$ into the expressions for the angles. $m\angle BGC=6x - 13=6\times8-13=48 - 13=35^{\circ}$, and $m\angle CGF=4x + 3=4\times8+3=32 + 3=35^{\circ}$.

Step5: Calculate $m\angle BGF$

$m\angle BGF=m\angle BGC+m\angle CGF$. So $m\angle BGF=35^{\circ}+35^{\circ}=70^{\circ}$.

Answer:

$70$