Question 243985
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Let *[tex \Large x] represent the measure of angle B.  Then *[tex \Large 3x] is the measure of angle A, and *[tex \Large 3x\ -\ 16] is the measure of angle C.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ x\ +\ (3x\ -\ 16)\ = 180]


Solve for *[tex \Large x] to get angle B.  Multiply the measure of B by 3 to get angle A.  Subtract 16 from the measure of A to get C.


Or, you could let *[tex \Large x] represent angle A.  Then *[tex \Large \frac{x}{3}] is the measure of angle B, and *[tex \Large x\ -\ 16] is the measure of angle C.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ \frac{x}{3}\ +\ (x\ -\ 16)\ = 180]


Or, you could let *[tex \Large x] represent angle C.  Then *[tex \Large x\ +\ 16] is the measure of angle A and *[tex \Large \frac{x+16}{3}] is the measure of angle B. So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 16)\ +\ \frac{x+16}{3}\ +\ x\ = 180]


Personally, I like the first way the best, but they all give the same answer in the end.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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