题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83008#problem/E
Description
HINT
题意：
给你一个水池，水池中放一个东西，上面一双眼睛，问眼睛能看到东西时，水面的最小高度
题解：
列列方程就是了
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** double w,h,x,xe,ye,p; int main()
{
int t=read();
while(t--)
{
cin>>w>>h>>x>>xe>>ye>>p;
double tan1=(xe-w)/(ye-h);
double tan2=tan(asin(sin(atan(tan1))/p)); double x1=w-tan1*h;
double ans=(x-x1)/(tan1-tan2);
if(ans<=)
cout<<"0.0000"<<endl;
else if(ans>h)
cout<<"Impossible"<<endl;
else
printf("%.4f\n",ans);
}
}