Question

for questions #10 - 14, refer to the diagram at the right, where ∠abc is a straight angle, m∠abf = 6x - 4, m∠fbe = 21x + 1, m∠erd = 12x - 8, and m∠dbc = 9x - 1.

1. find x.
2. find m∠fbe.
3. find m∠frd.
4. find the complement of ∠cbd in degrees.
5. find the supplement of ∠abf in degrees.

for questions #15 - 19, refer to the diagram at the right, where cd and fe intersect at g, m∠dgf = 7x + 9, and m∠cgf = 3x - 9.

Explanation:

Step1: Use the property of straight - angle

Since $\angle ABC$ is a straight angle, $m\angle ABF + m\angle FBE+m\angle EBD + m\angle DBC=180^{\circ}$. Given $m\angle ABF = 6x - 4$, $m\angle FBE=2x + 1$, $m\angle EBD = 12x-8$, and $m\angle DBC = 9x - 1$. Then $(6x - 4)+(2x + 1)+(12x-8)+(9x - 1)=180$.

Step2: Combine like - terms

Combine the $x$ terms and the constant terms: $(6x+2x + 12x+9x)+(-4 + 1-8 - 1)=180$. So $29x-12 = 180$.

Step3: Solve for $x$

Add 12 to both sides of the equation: $29x=180 + 12=192$. Then $x=\frac{192}{29}\approx6.62$.

Answer:

$x=\frac{192}{29}$