Question 177875
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For all triangles, the sum of measures of the interior angles is 180°.  So for your triangle *[tex \Large \angle A + \angle B + \angle C = 180]


But we are given that *[tex \Large \angle B = \frac {\angle A}{2} - 8] and


*[tex \Large \angle C = \angle A + 28].


We can substitute these expressions into the sum equation thus:


*[tex \Large \angle A + (\frac {\angle A}{2} - 8) + (\angle A + 28) = 180]


Now solve for *[tex \Large \angle A] by ordinary means.


Multiply by 2:


*[tex \Large 2 \angle A + \angle A - 16 + 2 \angle A + 56 = 360]


Collect like terms:


*[tex \Large 5 \angle A + 40 = 360]


Add -40 to both sides:


*[tex \Large 5 \angle A = 320]


Multiply by *[tex \Large \frac {1}{5}]


*[tex \Large \angle A = 64]


You should be able to figure the measures of the other two by substitution of this value.



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