Question

for each: geometry set - up/reason, substitute and solve. 21. geometry set - up. m∠aob = 5(x + 8)°, m∠aoc=(15x + 18)°, m∠boc = 9(x - 2)°. find m∠foe.

Explanation:

Step1: Assume the sum of angles around a point

The sum of angles around a point \(O\) is \(360^{\circ}\). So, \(m\angle AOB + m\angle AOC+m\angle BOC + m\angle FOE=360^{\circ}\). Substitute the given angle - expressions: \(5(x + 8)+(15x + 18)+9(x - 2)+m\angle FOE=360\).
First, expand the expressions:
\[

$$\begin{align*} 5x+40 + 15x+18+9x-18+m\angle FOE&=360\\ (5x + 15x+9x)+(40 + 18-18)+m\angle FOE&=360\\ 29x+40+m\angle FOE&=360 \end{align*}$$

\]
If we assume that \(\angle FOE\) is part of the full - circle sum and we want to solve for \(x\) first, we can rewrite the equation as \(5(x + 8)+(15x + 18)+9(x - 2)=360 - m\angle FOE\). Let's assume for now that we are only interested in the non - \(\angle FOE\) part to solve for \(x\).
\[

$$\begin{align*} 5(x + 8)+(15x + 18)+9(x - 2)&=360\\ 5x+40+15x + 18+9x-18&=360\\ (5x+15x + 9x)+(40 + 18-18)&=360\\ 29x+40&=360 \end{align*}$$

\]

Step2: Solve the equation for \(x\)

Subtract 40 from both sides of the equation \(29x+40 = 360\):
\[29x=360 - 40=320\]
Then \(x=\frac{320}{29}\approx11.03\)
Now, find \(m\angle FOE\):
\[

$$\begin{align*} m\angle FOE&=360-(5(x + 8)+(15x + 18)+9(x - 2))\\ &=360-(5x+40 + 15x+18+9x-18)\\ &=360-(29x + 40) \end{align*}$$

\]
Substitute \(x = \frac{320}{29}\) into the above formula:
\[

$$\begin{align*} m\angle FOE&=360-(29\times\frac{320}{29}+40)\\ &=360-(320 + 40)\\ &=360 - 360\\ &=0 \end{align*}$$

\]
This is wrong. Let's assume that \(\angle AOB+\angle AOC+\angle BOC = 180^{\circ}\) (a straight - line angle sum, assuming some of these angles form a straight line).
\[

$$\begin{align*} 5(x + 8)+(15x + 18)+9(x - 2)&=180\\ 5x+40+15x + 18+9x-18&=180\\ (5x+15x + 9x)+(40 + 18-18)&=180\\ 29x+40&=180 \end{align*}$$

\]
Subtract 40 from both sides: \(29x=180 - 40 = 140\), so \(x=\frac{140}{29}\approx4.83\)
\[

$$\begin{align*} m\angle FOE&=360-(5(x + 8)+(15x + 18)+9(x - 2))\\ &=360-(5x+40+15x + 18+9x-18)\\ &=360-(29x + 40) \end{align*}$$

\]
Substitute \(x=\frac{140}{29}\) into the formula for \(m\angle FOE\):
\[

$$\begin{align*} m\angle FOE&=360-(29\times\frac{140}{29}+40)\\ &=360-(140 + 40)\\ &=360 - 180\\ &=180 \end{align*}$$

\]
Equation: \(5(x + 8)+(15x + 18)+9(x - 2)=180\) (assuming \(\angle AOB+\angle AOC+\angle BOC\) form a straight - line)

Answer:

\(x=\frac{140}{29}\), \(m\angle FOE = 180^{\circ}\)