Question #01306

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Question #01306

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Answer 1 · 2017-12-20T18:52:18

$x = 30^{\circ}$ $x=30^circ$

Explanation:

I'm basing this solely on angle rules, not much else I can do for this. Someone may be able to provide a visual way of showing this.

Let's call the intersection $O$ $O$ .

We know that the angles in a triangle add up to $180^{\circ}$ $180^circ$ .

$∠ A E B = 180 - (25 + 55 + 45) = 180 - 125 = 55^{\circ}$ $/_AEB=180-(25+55+45)=180-125=55^circ$
$∠ A D B = 180 - (35 + 55 + 45) = 180 - 135 = 45^{\circ}$ $/_ADB=180-(35+55+45)=180-135=45^circ$
$∠ A O B = 180 - (55 + 45) = 180 - 100 = 80^{\circ}$ $/_AOB=180-(55+45)=180-100=80^circ$

Using the opposite angle rule, we know $∠ E O D = ∠ A O B$ $/_EOD=/_AOB$

$∠ O E D = 180 - (x + 80) = 180 - 80 - x = 100 - x$ $/_OED=180-(x+80)=180-80-x=100-x$

Since angles on a straight line add up to $180^{\circ}$ $180^circ$ :
$∠ D E F = 180 - (100 - x + 55) = x + 25$ $/_DEF=180-(100-x+55)=x+25$

$∠ A F B = 180 - (25 + 35 + 45 + 55) = 50^{\circ}$ $/_AFB=180-(25+35+45+55)=50^circ$

$∠ F D E = 180 - (50 + 25 + x) = 105 - x$ $/_FDE=180-(50+25+x)=105-x$

$∠ E D O = 180 - (45 + x + 105 - x) = 30 = x$ $/_EDO=180-(45+x+105-x)=30=x$

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Question #01306

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