Question

已知△ABC的三内角A,B,C所对三边分别为a,b,c，且sin(π/4+A)=7√2/10,0＜A＜π/4（1）求tanA的值（2）若

△ABC的面积S=24，b=8，求a的值

Answer 1

sin(π/4+A)=√2/2(sinA+cosA)=7√2/10;
(sinA+cosA)=7/5;
(sinA+cosA)=[2sin(A/2)cos(A/2)+cos(A/2)^2-sin(A/2)^2]/[sin(A/2)^2+cos(A/2)^2];
上下除以cos(A/2)^2得：
(sinA+cosA)=[2tan(A/2)+1-tan(A/2)^2]/[tan(A/2)^2+1]=7/5;
7tan(A/2)^2+7=10tan(A/2)+5-5tan(A/2)^2;
12tan(A/2)^2-10tan(A/2)+2=0;
6tan(A/2)^2-5tan(A/2)+1=0;
tan(A/2)=1/3或1/2;
tanA=[2tan(A/2)]/[1-tan(A/2)^2]=3/4或4/3;
而0＜A＜π/4；所以tanA<1;
所以tanA=3/4.
S=ab*tanA/2=24;
得：8*a*3/4*1/2=24;
a=8.