Whic are the two values of side AC for the triangle ABC with AB=9 cm, BC= 7cm and angleA=40? Hence, find angle C in each case.

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We'll apply cosine theorem, since we know the measure of an angle and the lengths of 2 sides:

BC^2 = AB^2 + AC^2 - 2AB*AC*cos 40

We'll substitute the values of the lengths known:

7^2 = 9^2 + AC^2 - 2*9*AC*cos 40

49 = 81 + + AC^2 - 13.78*AC

We'll get the quadratic:

AC^2 - 13.78*AC + 32 = 0

We'll apply quadratic formula:

AC = [13.78+sqrt(189.8884-128)]/2

AC = (13.78+7.86)/2

AC = 10.82 cm

AC = (13.78-7.86)/2

AC = 2.96 cm

**The 2 values of the lengths of the side AC are: AC=10.82 cm and AC = 2.96 cm.**

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