一道数学题，求速解，谢谢
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consider △BPC
BP/sin12°= BC/sin126°
BP = asin12°/sin54° (1)
consider △BPA
∠PAB = α
BP/sin∠PAB =AB/sin(180°-18° -α)
BP/sinα = a/sin(180°-18° -α)
BP = asinα/sin(162°-α) (2)
from (1) and (2)
asin12°/sin54° =asinα/sin(162°-α)
sin12°/sin54° =sinα/(sin162°.cosα - cos162°.sinα)
sin12°/sin54° =tanα/(sin162°- cos162°.tanα)
0.0794 +0.2444tanα =0.8090tanα
0.5646tanα=0.0794
α=8.005°
BP/sin12°= BC/sin126°
BP = asin12°/sin54° (1)
consider △BPA
∠PAB = α
BP/sin∠PAB =AB/sin(180°-18° -α)
BP/sinα = a/sin(180°-18° -α)
BP = asinα/sin(162°-α) (2)
from (1) and (2)
asin12°/sin54° =asinα/sin(162°-α)
sin12°/sin54° =sinα/(sin162°.cosα - cos162°.sinα)
sin12°/sin54° =tanα/(sin162°- cos162°.tanα)
0.0794 +0.2444tanα =0.8090tanα
0.5646tanα=0.0794
α=8.005°
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