所以我有一个三角形，点（100,90）距离（11）和线的角度（45°）我可以找到该线的另一个点吗？ 怎么样？

Question

okay so I am trying to draw a triangle, the triangle can be completely random, on a canvas in JavaScript

so I got the angles and the side for triangle ABC(this is not what I'm calling it in the code)

the sides

AB(11)
AC(12)
BC(13)

the angles which are solved in a function

BAC(69)
ABC(52)
BCA(59)

And the starting Point of the triangle at (100, 90)

The question I am having is how do I find the other points to the Triangle I thought the easiest way to draw it would be to draw a line that goes to each point

So I tired the mathematics with this code (I found on another page but )

function FindTriPoints(){
//Y2 = H(Sin(A)) + Y1
//X2 = Sqrt((H^2)-(Y2^2)) + X1
pointX1 = 100;
pointY1 = 90;

pointY2 = s3 * (Math.sin(angle1*Math.PI/180)) + pointY1;
pointX2 = Math.sqrt((s3 * s3) - (pointY2 * pointY2)) + pointX1;

alert("X2 = " + pointX2 + "\n Y2 = " + pointY2)
}

but X2 ends up becoming NaN because it is a negative value that it is trying to square root.

Edit Thanks to Cbroe and Jing3142 for helping me with Y2

Answer 1

well if you know valid triangle sides lengths (l1,l2,l3) and their angles (a,b,c) ...

• then it is quite simple with vector math ...

// compute directions
a1=0;
a2=180-b;
a3=a2+180-c;
a3=-b-c;
a3=-a;
// convert them from [deg] to [rad]
a1*=Math.pi/180.0;
a2*=Math.pi/180.0;
a3*=Math.pi/180.0;
// compute points
A=(x0,y0); // your start point is known
B=A+l1*dir(a0)=(x0+l1*Math.cos(a0),y0+l1*Math.sin(a0));
B=A+l1*dir( 0)=(x0+l1               ,y0                 );     // a0 is always zero
C=A-l3*dir(a3)=(x0-l3*Math.cos(a3),y0-l3*Math.sin(a3));    // C from A point
C=B+l2*dir(a2)=(x0+l1+l2*Math.cos(a2),y0+l2*Math.sin(a2)); // C from B point

[notes]

• as you can see there are more alternatives for some variables choose one you like
• do not forget to check if l1+l2>l3 and l1+l3>l2 and l2+l3>l1
• if not then your lengths are not valid triangle sides
• also a+b+c = 180
• if not then your angle computation is wrong
• if you want to over-check the triangle then compute C from A and from B point
• if they are not the same (their distance > some accuracy constant like 0.001)
• then it is not a valid triangle