在ΔABC中,c=√2+√6,∠C=30°,求a+b的取值范围.

2个回答

  • 由正弦定理得:a/sinA=b/sinB=c/sinC=2R (R为△ABC 外接圆的半径)

    则:a=2R sinA,b=2R sinB ,

    2R = c/sinC = (√2+√6)/ sin30° = 2(√2+√6)

    ∵ C=30°,A + B = 150°

    ∴ A=150°-B,且 0°<B<150°

    ∴ a+b = 2R (sinA + sinB)

    = 2R [sin(150°-B) + sinB] ( 注:sin(150°-B)=sin150°cosB - cos150°sinB )

    = (√2+√6) × [cosB+(√3+2)sinB]

    = [(√2+√6)的平方] × [(√6--√2)/4?cosB + (√6--√2)/4?sinB]

    = (8+4√3) × sin(15°+B)

    ∵ 0° < B < 150°

    ∴15°< (B+15°) < 165°

    ∴ sin15° < sin(B+15°) ≤1

    ∴ [(√6-√2)/4] < sin(B+15°) ≤ 1

    ∴ [(√6-√2)/4] ×(8+4√3) < [(8+4√3) × sin(15°+B)] ≤ (8+4√3)

    ∴ (√2+√6)<(a+b) ≤(8+4√3)

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