median length x = BC length
median hits BC at D
triangle ABC
4^2 = 3^2 + x^2 - 6 x cos B
triangle ABD
x^2 = 3^2 + (x/2)^2 - 3 x cos B
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16 = 9 + x^2 - 6x cos B
x^2 = 9 +x^2/4 -3 x cos B
x^2 -6 x cos B -7 = 0
3x^2/4 +3 x cos B - 9 = 0
x^2 - 6 x cos B - 7 = 0
(3/2)x^2 + 6 x cos B -18 = 0
(5/2)x^2 - 25 = 0
5 x^2 = 50
x^2 = 10
x = sqrt(10)
In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC have the same length. What is BC?
Not making sense to me, I think the answer must be simple, but I don't know how to solve I applied the law of sines but to no avail. Help is appreciated, thanks.
2 answers
thanks a lot man!