在三角形ABC中，内角A、B、C所对边为a,b,c,已知a=1,b=2，cosC=1 在三角形ABC中，内角A，B，C所对的边分别为a，b，c，已...
\u5728\u25b3ABC\u4e2d\uff0c\u5185\u89d2A,B,C\u6240\u5bf9\u7684\u8fb9\u5206\u522b\u4e3aa\uff0cb\uff0cc\uff0c\u5df2\u77e5A=\u03c0/4\uff0cb²-a²=c²/2. \uff081tanC\u7684\u503c\u89e3\u6cd5\u5982\u4e0b\uff1a
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\uff081\uff09\u2235a=bcosC+csinB\uff0c\u2234\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0c\u5f97sinA=sinBcosC+sinBsinC\u2026\u2460\uff0c\u53c8\u2235sinA=sin\uff08B+C\uff09=sinBcosC+cosBsinC\u2026\u2461\uff0c\u2234\u6bd4\u8f83\u2460\u2461\uff0c\u53ef\u5f97sinB=cosB\uff0c\u5373tanB=1\uff0c\u7ed3\u5408B\u4e3a\u4e09\u89d2\u5f62\u7684\u5185\u89d2\uff0c\u53ef\u5f97B=45\u00b0\uff1b\uff082\uff09\u2235\u25b3ABC\u4e2d\uff0cb=2\uff0cB=45\u00b0\uff0c\u2234\u6839\u636e\u4f59\u5f26\u5b9a\u7406b2=a2+c2-2accosB\uff0c\u53ef\u5f97a2+c2-2accos45\u00b0=4\uff0c\u5316\u7b80\u53ef\u5f97a2+c2-2ac=4\uff0c\u2235a2+c2\u22652ac\uff0c\u22344=a2+c2-2ac\u2265\uff082-2\uff09ac\uff0e\u7531\u6b64\u53ef\u5f97ac\u226442-2=4+22\uff0c\u5f53\u4e14\u4ec5\u5f53a=c\u65f6\u7b49\u53f7\u6210\u7acb\uff0e\u2234\u25b3ABC\u9762\u79efS=12acsinB=24ac\u226424\uff084+22\uff09=2+1\uff0e\u7efc\u4e0a\u6240\u8ff0\uff0c\u5f53\u4e14\u4ec5\u5f53a=c\u65f6\uff0c\u25b3ABC\u9762\u79efS\u7684\u6700\u5927\u503c\u4e3a2+1\uff0e
是求c吗c²=a²+b²-2abcosC
=1²+2²-2×1×2×1
=1+4-4
=1
c=1
解：因为cosC=1，所以∠C=90°，求边用勾股定理、求角用角与边的函数关系即可求得。
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