在锐角三角形ABC中，已知AB＝8，AC＝10，角ABC的面积为25，求角A的各个三角函数值
\u5728\u9510\u89d2\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u5df2\u77e5AB=8\uff0cAC=10\uff0c\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef\u4e3a25\uff0c\u6c42\u2220A\u7684\u5404\u4e2a\u4e09\u89d2\u51fd\u6570\u503c\u56e0\u4e3a\u4e09\u89d2\u5f62ABC\u7684\u9762\u79ef\u4e3a25\uff0c\u8bbeAB\u8fb9\u4e0a\u7684\u9ad8\u4e3ah\u3002\u6709
8h/2=25
h=25/4
tanA=h/AB=25/4/8=25/32
\u53d6\u53c2\u8003\u5706\uff0cr = \u221a(32²+25²) = 40.61
sinA=25/40.61 =0.6156
cosA=32/40.61 =0.7882
S\u0394=1/2AB*BC*sinA=40sinA=32\uff0c
sinA=4/5\uff0c\u2235\u0394ABC\u662f\u9510\u89d2\u4e09\u89d2\u5f62\uff0c
\u2234cosA=3/5\uff0c
\u2234AC^2=AB^2+BC^2-2AB*BC*cosA
=100+64-96
=68\uff0c
\u2234AC=2\u221a17\u3002
∵AB=8，AC=10，S=25
∴sinA=5/8
又∵A为锐角
∴cosA=√（1-sin²A）=√39/8
tanA=sinA/cosA=5√39/39
cotA=1/tanA=√39/5
正弦是8分之5，余弦8分之根号炸39，正切根号下39分之5
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