如图,在△ABC中,∠B=45°,∠A=15°,BC=√3-1,求AC,AB的长.
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如图,在△ABC中,∠B=45°,∠A=15°,BC=√3-1,求AC,AB的长. (图2)如图,过a点做ae⊥于bc,交bc的延长线于e点,垂足为e.
依题意有:be = ae;∠cae = 45° -15° = 30°
∴ ce = ac÷2;ae=√3ac÷2
故有:bc + ce = √3 -1 + ac÷2 = √3ac÷2
即:(√3 -1 )×2 = (√3 -1)ac解得:ac = 2
∴ ce = ac÷2 = 1,则be = bc+ce = √3 =ae
∴ ab = √(be² +ae²) = √6