Question 237386
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I presume that you mean B to be the angle that is different in measure from the other two angles in the isosceles triangle.  The word vertex is used incorrectly in this context.  All triangles, regardless of their configuration have three vertices.  In an isosceles triangle, two of the vertices have angles that are equal in measure.


Given a correct assumption on my part, if B is the unequal angle, then angle A must be congruent to angle C, hence their measures are equal.  Therefore if angle A measures *[tex \Large 3x\ +\ 2] then angle C must measure *[tex \Large 3x\ +\ 2] also.


We know that the sum of the measures of the interior angles of a triangle is 180 degrees or *[tex \Large \pi] radians.


So we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (8x\ -\ 6)\ +\ (3x\ +\ 2)\ +\ (3x\ +\ 2)\ =\ 180\ \Rightarrow\ 14x\ -\ 2\ =\ 180\ \Rightarrow\ 14x\ -\ 2\ =\ 180\ \Rightarrow\ x\ =\ 13]


From which you can calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ B\ =\ 8(13)\ -\ 6\ =\ 98] degrees.


and 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ C\ =\ 3(13)\ +\ 2\ =\ 41] degrees.


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 98\ +\ 41\ + 41\ =\ 180]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (8x\ -\ 6)\ +\ (3x\ +\ 2)\ +\ (3x\ +\ 2)\ =\ \pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14x\ -\ 2\ =\ \pi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\pi\ +\ 2}{14]


From which you can calculate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ B\ =\ \frac{4\pi\ +\ 8}{7}\ -\ 6\ =\ \frac{4\pi\ -\ 34}{7}] radians


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ C\ =\ \frac{3\pi\ +\ 6}{14}\ + 2\ =\ \frac{3\pi\ +\ 34}{14}] radians


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4\pi\ -\ 34}{7}\ +\ \frac{3\pi\ +\ 34}{14}\ +\ \frac{3\pi\ +\ 34}{14}\ =\ \pi]


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