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  • 已知三角形ABC的周長(zhǎng)為根號(hào)2+1,且sinA+sinB=根號(hào)2乘以sinC

    已知三角形ABC的周長(zhǎng)為根號(hào)2+1,且sinA+sinB=根號(hào)2乘以sinC
    (1)求邊AB的長(zhǎng) (2)若三角形ABC的面積為六分之一sinC,求角C的度數(shù)
    數(shù)學(xué)人氣:385 ℃時(shí)間:2019-08-20 17:33:04
    優(yōu)質(zhì)解答
    sinA + sinB = √2sinC
    a/sinA = b/sinB = c/sinC
    有:
    (a+b+c)/(sinA+sinB+sinC) =(a+b)/(sinA+sinB)=c/sinC
    所以有:
    (√2 + 1 - c)/(√2sinC) = c/sinC
    那么:
    √2 + 1 - c = √2c
    所以,AB = c = 1
    S = 1/2 * a * b * sinC = 1/6 sinC
    所以,a * b = 1/3.
    又 a + b = (√2 + 1) - c = √2
    因此,可以得到:
    (√2 - a) * a = √2*a - a^2 = 1/3
    化簡(jiǎn),得到:
    3a^2 - 3√2 * a + 1 = 0
    a = (3√2 - √6)/6 = √2/2 - √6/6,b = √2/2 + √6/6
    或者,a = (3√2 + √6)/6 = √2/2 + √6/6,b = √2/2 - √6/6
    又因?yàn)椋篶osC = (c^2 - a^2 - b^2)/(2ab)
    = [1 - (1/2 + √12/6 + 1/6) - (1/2 - √12/6 + 1/6)]/[2 * (1/2 - 1/6)]
    = [-1/3]/(2/3)
    = -1/2
    所以,C = 120°
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