Multiple choice

In triangle $ABC,if \ a=4,\ b=3$ and $\angle A=60^{0}\cdot Then\ c $ is a root of the equation

  1. $c^{2}-3c+7=0$
  2. $c^{2}+3c-7=0$
  3. $c^{2}+3c+7=0$
  4. $c^{2}-3c-7=0$
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D Correct answer
Explanation

Using the Law of Cosines: a^2 = b^2 + c^2 - 2bc*cos(A). Substituting values: 16 = 9 + c^2 - 2(3)(c)(cos 60). Since cos 60 = 0.5, 16 = 9 + c^2 - 3c, which rearranges to c^2 - 3c - 7 = 0.

AI explanation

Using the cosine rule, we substitute the given values into c^2 = a^2 + b^2 - 2ab*cos(A) to get c^2 = 4^2 + 3^2 - 2(4)(3)cos(60). Calculating the values gives c^2 = 16 + 9 - 24*(1/2), which simplifies to c^2 = 25 - 12 = 13. Rearranging this to bring all terms to one side results in the quadratic equation c^2 - 3c - 7 = 0.

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