Question

for each: geometry set - up/reason, substitute and solve. 21. 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 is 360°

$m\angle AOB + m\angle BOC+m\angle AOC + m\angle FOE=360^{\circ}$

Step2: Substitute the given angle - expressions

$5(x + 8)+9(x - 2)+(15x + 18)+m\angle FOE=360$

Step3: Expand the expressions

$5x+40 + 9x-18+15x + 18+m\angle FOE=360$

Step4: Combine like - terms

$(5x+9x + 15x)+(40-18 + 18)+m\angle FOE=360$
$29x+40+m\angle FOE=360$

Step5: Isolate $m\angle FOE$

$m\angle FOE=360-(29x + 40)$
$m\angle FOE=320-29x$

However, if we assume that the non - given angles between these angles are 0 (or there is some additional information about the relationship of these angles that makes the sum of these three angles equal to 360°), we first solve for $x$ from the sum of the three given angles:

$5(x + 8)+9(x - 2)+(15x + 18)=360$
$5x+40+9x - 18+15x + 18=360$
$29x+40=360$
$29x=360 - 40$
$29x=320$
$x=\frac{320}{29}$

Then:
$m\angle AOB=5(x + 8)=5(\frac{320}{29}+8)=5(\frac{320+232}{29})=5\times\frac{552}{29}=\frac{2760}{29}\approx95.17^{\circ}$
$m\angle BOC=9(x - 2)=9(\frac{320}{29}-2)=9(\frac{320 - 58}{29})=9\times\frac{262}{29}=\frac{2358}{29}\approx81.31^{\circ}$
$m\angle AOC=15x + 18=15\times\frac{320}{29}+18=\frac{4800}{29}+18=\frac{4800+522}{29}=\frac{5322}{29}\approx183.52^{\circ}$
$m\angle FOE=360-(m\angle AOB + m\angle BOC+m\angle AOC)$
$m\angle FOE=360-(\frac{2760}{29}+\frac{2358}{29}+\frac{5322}{29})=360-\frac{2760 + 2358+5322}{29}=360-\frac{10440}{29}=\frac{10440 - 10440}{29}=0^{\circ}$ (assuming no other non - accounted angles)

If we assume these angles are part of a non - full 360° set of angles around a point and there is no other information about the relationship of angles, we leave $m\angle FOE = 320-29x$

If we assume the sum of these three given angles is 360° (a wrong assumption if there are other non - shown angles in the full 360° around the point, but for the sake of solving just with these three angles):

Answer:

If we solve for $x$ from $5(x + 8)+9(x - 2)+(15x + 18)=360$:
$x=\frac{320}{29}$
$m\angle AOB=\frac{2760}{29}\approx95.17^{\circ}$
$m\angle BOC=\frac{2358}{29}\approx81.31^{\circ}$
$m\angle AOC=\frac{5322}{29}\approx183.52^{\circ}$
$m\angle FOE = 0^{\circ}$ (assuming no other non - accounted angles)

If we consider the general case with the sum of all angles around a point being 360° and no other information about non - given angles:
$m\angle FOE=320-29x$