Question

in the figure, ca and ce are opposite rays, ch bisects ∠gcd, and gc bisects ∠bgd. if m∠bgc=(7x - 2)° and m∠cgf=(8x - 8)°, what is m∠bgf? m∠bgf =

Explanation:

Step1: Use angle - addition postulate

Since $\overrightarrow{CA}$ and $\overrightarrow{CE}$ are opposite rays, $\angle BGC+\angle CGF = 180^{\circ}$ (linear - pair of angles). So, $(7x - 2)+(8x - 8)=180$.

Step2: Simplify the equation

Combine like - terms: $7x+8x-2 - 8 = 180$, which gives $15x-10 = 180$.

Step3: Solve for x

Add 10 to both sides of the equation: $15x=180 + 10=190$. Then $x=\frac{190}{15}=\frac{38}{3}$.

Step4: Find $\angle BGF$

We know that $\angle BGF=\angle BGC+\angle CGF$. Substitute $x = \frac{38}{3}$ into the expressions for $\angle BGC$ and $\angle CGF$.
$\angle BGC=7x - 2=7\times\frac{38}{3}-2=\frac{266}{3}-2=\frac{266 - 6}{3}=\frac{260}{3}$ and $\angle CGF=8x - 8=8\times\frac{38}{3}-8=\frac{304}{3}-8=\frac{304 - 24}{3}=\frac{280}{3}$.
$\angle BGF=(7x - 2)+(8x - 8)=15x-10$. Substitute $x=\frac{38}{3}$ into $15x - 10$: $15\times\frac{38}{3}-10=190 - 10 = 180$.

Answer:

$180$