【答案】
分析:本題可以利用銳角三角函數(shù)的定義求解,也可以利用互為余角的三角函數(shù)關(guān)系式求解.
解答:解:解法1:利用三角函數(shù)的定義及勾股定理求解.
∵在Rt△ABC中,∠C=90°,
∴sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/0.png)
,tanB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/1.png)
和a
2+b
2=c
2.
∵sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/2.png)
,設(shè)a=3x,則c=5x,結(jié)合a
2+b
2=c
2得b=4x.
∴tanB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/3.png)
.
故選A.
解法2:利用同角、互為余角的三角函數(shù)關(guān)系式求解.
∵A、B互為余角,
∴cosB=sin(90°-B)=sinA=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/4.png)
.
又∵sin
2B+cos
2B=1,
∴sinB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/6.png)
,
∴tanB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/8.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022155318307402008/SYS201310221553183074020001_DA/9.png)
.
故選A.
點(diǎn)評:求銳角的三角函數(shù)值的方法:利用銳角三角函數(shù)的定義,通過設(shè)參數(shù)的方法求三角函數(shù)值,或者利用同角(或余角)的三角函數(shù)關(guān)系式求三角函數(shù)值.