在三角形ABC中,内角A.B.C的对边分别为a.b.c,且bsinA=根号3acosB求角B的大小？
sinA = √3a cosBb/a = √3 cosB/sinA
根据正弦定理：b/a = sinB/sinA
∴sinB/sinA = √3 cosB/sinA
sinB = √3 cosB
tanB = √3
B=π/6,7,
绛旓細B=蟺/6,7,
绛旓細鍦ㄤ笁瑙掑舰涓紝鏈夈愭寮﹀畾鐞嗐戯細asinB=bsinA.鎵浠ワ紝bsinA=鏍瑰彿3acosB锛屽彲浠ュ寲涓 asinB=鏍瑰彿3acosB锛a涓嶆槸0锛屽悓闄や互a锛屽緱鍒 sinB = 鏍瑰彿3 cosB锛屽綋B涓虹洿瑙掓椂锛屽彸杈逛负0锛屽乏杈逛负1锛屼笉绛夈傛墍浠涓嶆槸鐩磋锛宑osB涓嶄负0锛屽悓闄や互cosB寰楀埌 tanB = 鏍瑰彿3. B=60搴︺
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