设球o半径为1,A,B,C是球面上三点，已知A到B,C两点的球面距离都是π/2,且二面角B-OA-C的大小为π/3,则
\u8bbe\u7403O\u7684\u534a\u5f84\u662f1\uff0cA\u3001B\u3001C\u662f\u7403\u9762\u4e0a\u4e09\u70b9\uff0c\u5df2\u77e5A\u5230B\u3001C\u4e24\u70b9\u7684\u7403\u9762\u8ddd\u79bb\u90fd\u662f \uff0c\u4e14\u4e8c\u9762\u89d2B-OA-C\u7684\u5927\u5c0f\u662f \uff0c\u5219C
OA\uff0cOB\uff0cOC
\u4e24\u4e24\u5782\u76f4
\uff08\u60f3\u60f3\u6b63\u65b9\u4f53\uff09
\u7b54\u6848\uff08\u6839\u53f73\uff09/3
同理,角AOC=π/2=90度(AO与CO垂直).
所以AO与平面BOC垂直,所以角BOC度数=二面角B-OA-C度数=π/3.
所以B与C的球面距离为π/3，
所求最短距离即球面距离为π/2+π/2+π/3=4π/3
就是求球面三角形ABC的周长,关键是求BC,
因为半径是1,故大圆周长是2Pi,(1/4)周长是Pi/2,又因为A到B,C两点的球面距离都是π/2,故如果把A点看作北极点,则,B,C两点在赤道上,
故球面三角形ABC的周长=(2*Pi)/4+2*(Pi/2)=3Pi/2.
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绛旓細鐞冨崐寰R=1 瑙扐OB=蟺/2,AB=2sin(蟺/4)=鏍瑰彿2,AC=AB=鏍瑰彿2 瑙BOC=蟺/3,BC=2sin(蟺/6)=1 sin(瑙払AC/2)=(1/2)/(鏍瑰彿2)=1/(2(鏍瑰彿2))=(鏍瑰彿2)/4 cos(瑙払AC/2)=(1-((鏍瑰彿2)/4)^2)^(1/2)=(1/4)(鏍瑰彿(14)sin(瑙払AC)=2sin(瑙払AC/2)cos(瑙払AC/2)=(1/4)...
