1.
S△ABC = AB * AC * sin角BAC / 2 = 6 * 4 * sin120° / 2 = 6倍根號(hào)3
2.
根據(jù)余弦定理
BC^2 = AB^2 + AC^2 - 2*AB*AC*cos角BAC = 36 + 16 + 2 * 6 * 4 * 1/2 = 36 + 16 + 24 = 76
BC=2倍根號(hào)19
3.
S△ABC = BA * BC * sin角B/2 = 3 * 2倍根號(hào)19 * sin角B
所以6倍根號(hào)19 * sinB = 6倍根號(hào)3,得sinB = 根號(hào)3 / 根號(hào)19
根據(jù)余弦定理
AC^2 = BA^2 + BC^2 - 2 * BA * BC * cosB
16 = 36 + 76 - 2 * 6 * 2倍根號(hào)19 * cosB
根號(hào)19 * cosB = 4,得cosB = 4/根號(hào)19
所以tanB = sinB / cosB = (根號(hào)3/根號(hào)19) / (4/根號(hào)19) = 根號(hào)3/4第一問(wèn)怎么出來(lái)的啊。。。為什么△的面積等于那個(gè)公式？