Question 108135: In a triangle ABC,angle B is 3 times angle A and angle C is 19 degrees less than 6 times angle A. Find the size of the angles.
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! From the problem you can see that all three angles are expressed in terms of angle A.
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First there is angle A.
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Then there is angle B which you are told is three times angle A ... or 3A
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Finally, there is angle C which is six times angle A ... take away 19 degrees ... which is 6A - 19
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But the sum of all three angles has to equal 180 degrees for this triangle. So you can add
the three angles and set the sum equal to 180:
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A + 3A + 6A - 19 = 180
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Get rid of the -19 on the left side of the equation by adding 19 to both sides to get:
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A + 3A + 6A = 199
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Now add up all the terms on the left side and the equation reduces to:
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10A = 199
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Solve for angle A by dividing both sides by 10. When you do that division, you get that
angle A = 19.9 degrees.
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Since you know that angle B must be 3 times angle A you can multiply angle A by 3 (19.9 * 3)
to find that angle B = 59.7 degrees. Finally, you can find angle C by multiplying
angle A by 6 and subtracting 19 degrees. Six times angle A is 6 * 19.9 and this equals 119.4 degrees
and then subtracting 19 degrees results in 100.4 degrees.
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In summary, the measures of the three angles of this triangle are angle A = 19.9 degrees,
angle B = 59.7 degrees, and angle C = 100.4 degrees.
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If you add them the result is 180 degrees, just as it should be.
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Hope this helps you to understand the problem and how to solve it.
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