How do you find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third angle measures 3 times the second angle decreased by 12?

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How do you find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third angle measures 3 times the second angle decreased by 12?

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Answer 1 · 2018-05-09T10:01:29

$∠ A = 32^{\circ}, ∠ B = 64^{\circ}, ∠ C = 84^{\circ}$ $/_A=32^@, /_B=64^@, /_C=84^@$

Explanation:

To be certain I understand your question correctly, you have a triangle, where you with "measures" mean number of degrees each angle is.

I.e. let the triangle look something like this:

enter image source here

If $∠ A = a^{\circ}$ $/_A=a^@$
Then $∠ B = 2 ∠ A = 2 a^{\circ}$ $/_B=2/_A=2a^@$
And $∠ C = 3 ∠ A - 12^{\circ} = 3 a^{\circ} - 12^{\circ}$ $/_C=3/_A-12^@=3a^@-12^@$

If my understanding of your question is correct, we have
$∠ A + ∠ B + ∠ C = 180^{\circ}$ $/_A+/_B+/_C=180^@$
Therefore $a^{\circ} + 2 a^{\circ} + 3 a^{\circ} - 12^{\circ} = 180^{\circ}$ $a^@+2a^@+3a^@-12^@=180^@$

This gives $6 a^{\circ} - 12^{\circ} = 180^{\circ}$ $6a^@-12^@=180^@$
Or $a^{\circ} = 32^{\circ}$ $a^@=32^@$

The angles in the triangle, therefore, are
I.e. $∠ A = a^{\circ} = 32^{\circ}$ $/_A=a^@=32^@$
Then $∠ B = 2 ∠ A = 2 a^{\circ} = 64^{\circ}$ $/_B=2/_A=2a^@=64^@$
And $∠ C = 3 ∠ A - 12^{\circ} = 84^{\circ}$ $/_C=3/_A-12^@=84^@$

This gives the following triangle:

enter image source here

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How do you find the measures of the angles of a triangle if the measure of one angle is twice the measure of a second angle and the third angle measures 3 times the second angle decreased by 12?

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