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What is the measure of AC ? Enter your answer in the box. ° Circle with inscribed angle A B C. Angle A B C is 3 x minus 1.5 degrees. The intercepted arc A C is 3 x plus 9 degrees.

Respuesta :

Inscribed angles are 1/2 of the measure of the intercepted arc.  That means from this we have the equation
[tex](3x-1.5)=\frac{1}{2}(3x+9)[/tex].  Distribute the 1/2:
[tex]3x-1.5=\frac{1}{2}*3x+\frac{1}{2}*9 \\3x-1.5=\frac{1}{2}*\frac{3x}{1} + \frac{1}{2}*\frac{9}{1} \\3x-1.5=\frac{3x}{2}+\frac{9}{2} \\3x-1.5=1.5x+4.5[/tex]
We cannot have a variable on both sides of the equation, so cancel the 1.5x to avoid negatives:
[tex]3x-1.5-1.5x=1.5x+4.5-1.5x[/tex]
Combine like terms:
[tex]1.5x-1.5=4.5[/tex]
Cancel the -1.5 by adding:
[tex]1.5x-1.5+1.5=4.5+1.5 \\1.5x=6[/tex]
Divide both sides by 1.5:
[tex]\frac{1.5x}{1.5}=\frac{6}{1.5} \\x=4[/tex]
Since AC is 3x+9, we substitute 4 in for x:
3*4+9=12+9=21°
Ver imagen MsEHolt
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Use the Inscribed Angle theorem to get the measure of AC. The intercepted arc AC is [tex]21^{\circ}[/tex].

Given,

The inscribed angle [tex]\angle ABC[/tex] is [tex]3x-1.5[/tex].

And the Intercepted arc AC is [tex]3x+9[/tex].

What is the Inscribed Angle theorem?

We know that, Inscribed Angle Theorem stated that the measure of an inscribed angle is half the measure of the intercepted arc.

So,

[tex]3x-1.5=\frac{1}{2} (3x+9)[/tex]

[tex]2 (3x-1.5)=3x+9[/tex]

[tex]6x-3.0=3x+9[/tex]

[tex]6x-3x=9+3[/tex]

[tex]3x=12\\x=4[/tex]

Since the intercepted arc AC is [tex]3x+9[/tex], putting the value of [tex]x=4[/tex] we get,

intercepted arc AC is [tex]21^{\circ}[/tex].

Hence the intercepted arc AC is [tex]21^{\circ}[/tex].

For more details on the Inscribed Angle theorem follow the link:

https://brainly.com/question/5436956

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