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已知(a+b+c)*(b+c-c)=3bc,sinA=2sinBsinC.求三角形ABC的形状?我太笨了
人气:450 ℃ 时间:2020-05-24 01:04:29
解答
(A + B + C)(B + CA)= 3BC
[(B + C)+ A] [(B + C)-α] = 3BC
(B + C)^ 2 - 一个^ 2 = 3BC
B ^ 2 +2 BC + C ^ 2-A ^ 2 = 3BC
B ^ 2-BC + C ^ 2 = A ^ 2
是有规律的余弦^ 2 = B ^ 2 + C ^ 2-2bccosB
B ^ 2-BC + C ^ 2 = B ^ 2 + C ^ 2-2bccosB
BC = 2bccosB
COSB = 1/2
B = 60度

= 2sinBcosC
=√3cosC
= - √3cos(A + B)
= - √3cos(A +60)
= - √3(cosAcos60-sinAsin60)
= - √3(cosA/2-√3sinA / 2)
= - √3cosA / 2新浪+3 / 2
新浪=√3cosA
A = 60°
C = 180-60-60 = 60度

等边三角形
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