三角形ABC,AB=2.AC=4.角BAC=120度,D是BC中点,求AD=?
1个回答
由余弦定理：
a^2 = b^2+ c^2 -2*bc*cos(A);
|BC|^2 = 2^2 + 1^2 -2*2*1*cos(120) = 7;
|BC| = sqrt(7);
|DC| = 2|BD|,|DC| = 2*sqrt(7)/3,|BD| = sqrt(7)/3;
考虑三角形ABD;
cos(角ABC) = (|AB|^2+|BC|^2 - |AC|^2)|/(2|AB|*|BC|)
= (4+7-1)/(2*2*sqrt(7))
= 5*sqrt(7)/14;
|AD|^2 = |AB|^2 +|BD|^2 -2*|AB|*|BD|*cos(角ABD)
= 13/9;
cos(角ADC) = (|AD|^2 +|DC|^2 -|AC|^2)/(2*|AD|*|DC|)
= 8*sqrt(91)/91;
由向量点乘法则：
向量AD.向量BC = |AD|*|BC|*cos(向量AD与向量BC的夹角)
= (sqrt(13)/3) * sqrt(7)*(-8*sqrt(91)/91)
= -8/3.
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