半径为1的球面上有A、B、C三点，其中点A与B、C两点间的球面距离均为π2，且B、C两点间的球面距离为π3，

A\u3001B\u3001C\u662f\u534a\u5f84\u4e3a1\u7684\u7403\u9762\u4e0a\u4e09\u70b9\uff0cB\u3001C\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a \u03c0 3 \uff0c\u70b9A\u4e0eB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a

\uff081\uff09\u2235\u7403\u9762\u8ddd\u79bb?=\u03b8?r\uff08\u03b8\u4e3a\u52a3\u5f27\u6240\u5bf9\u5706\u5fc3\u89d2\uff09\uff0c\u4e14B\u3001C\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a \u03c0 3 \uff0c\u70b9A\u4e0eB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a \u03c0 2 \uff0c\u6545\u5f97\u2220AOB= \u03c0 2 \uff0c\u2220BOC= \u03c0 3 \uff0c\u2220AOC= \u03c0 2 \uff1b\uff082\uff09\u2235OA=OB=OC=1\uff0c\u2234AB=AC= 2 \uff0cBC=1\uff0c\u2234S \u25b3OBC = 3 4 \uff0cS \u25b3ABC = 7 4 V 0 -ABC= 1 3 ? 3 4 ?1= 1 3 ? 7 4 ?d\uff0c\u2234d= 21 7 \uff0c\u7403\u5fc3\u5230\u622a\u9762ABC\u7684\u8ddd\u79bb\u4e3a 21 7 \uff0c\uff083\uff09\u8bbe\u7403\u7684\u5185\u63a5\u6b63\u65b9\u4f53\u68f1\u957f\u4e3aa\uff0c\u6839\u636e\u7403\u7684\u76f4\u5f84\u4e3a\u6b63\u65b9\u4f53\u7684\u5bf9\u89d2\u7ebf\uff0c\u5219 3 a=2\uff0c\u2234a= 2 3 3 \uff0c\u2234S \u6b63\u65b9\u4f53 \uff1aS \u7403\u9762 =6? ( 2 3 3 ) 2 \uff1a4\u041b=2\uff1a\u041b\uff0e

\u89e3\uff1a\u5982\u56fe\uff0c\uff081\uff09\u56e0\u4e3a\u7403O\u7684\u534a\u5f84\u4e3a1\uff0cB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a\u03c03\uff0c\u70b9A\u4e0eB\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a\u03c02\uff0c\u6240\u4ee5\u2220BOC=\u03c03\uff0c\u2220AOB=\u2220AOC=\u03c02\uff0e\uff082\uff09\u56e0\u4e3aBC=1\uff0cAC=AB=2\uff0c\u6240\u4ee5\u7531\u4f59\u5f26\u5b9a\u7406\u5f97cos\u2220BAC=34\uff0csin\u2220BAC=74\uff0c\u8bbe\u622a\u9762\u5706\u7684\u5706\u5fc3\u4e3aO1\uff0c\u8fde\u63a5AO1\uff0c\u5219\u622a\u9762\u5706\u7684\u534a\u5f84R=AO1\uff0c\u7531\u6b63\u5f26\u5b9a\u7406\u5f97r=BC2sin\u2220BAC=277\uff0c\u6240\u4ee5OO1=OA2?r2=217\uff0e

解：球心O与A，B，C三点构成三棱锥O-ABC，如图所示，
已知OA=OB=OC=R=1，∠AOB=∠AOC=90°，∠BOC=60°，
由此可得AO⊥面BOC．
∵S△BOC＝

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