A circle inscribed in a triangle ABC touches the side AB at D such that AD = 5 and BD = 3. If $\displaystyle \angle A= 60^{\circ},$ then the value of $\displaystyle \left [ BC/3 \right ]$ (where $\displaystyle \left [\cdot \right ]$ represents the greatest integer function) is ________ .
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$4$
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$5$
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$6$
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None of these
A
Correct answer
Explanation
In a triangle with an incircle, AD = AF = 5 and BD = BE = 3. Let CE = CF = x. Sides are c=8, a=3+x, b=5+x. Using the Law of Cosines: a^2 = b^2 + c^2 - 2bc cos(60). (3+x)^2 = (5+x)^2 + 8^2 - 2(5+x)(8)(0.5). 9 + 6x + x^2 = 25 + 10x + x^2 + 64 - 40 - 8x. 9 + 6x = 49 + 2x. 4x = 40, so x = 10. BC = a = 3+10 = 13. BC/3 = 13/3 = 4.33. The greatest integer function [4.33] = 4.