Plotted 30-60-90 Triangle, Find Coordinate of Last Point

[iuhn]

Hello! So as #26 says on the attached photo, "△ABC is a 30°-60°-90° triangle", and I have to find the coordinates of A. I know that the hypotenuse is 6 units long, making the shorter leg 3 units long and the longer leg 3√3 or approximately 5.20 units long, but I don't know how to use that to find the coordinates of A. My teacher said something about the distance formula. I've tried plugging it in and using a system of equations to solve it since there are two variables, x and y (the coordinates of A), and but I gave up because of how complex it is, leading me to believe that I was doing it wrong. Would someone pretty please explain to me how to go about this?

 

[MarkFL]

Hello, and welcome to FMH!

I would also use the distance formula, in the manner you are attempting. We know $\overline{AB}=3$ and $\overline{AC}=3\sqrt{3}$, and so, letting the coordinates of point $A$ be $(x,y)$ we may write:

[MATH](x-3)^2+(y+1)^2=9[/MATH]
[MATH](x+3)^2+(y+1)^2=27[/MATH]
Notice that:

[MATH](y+1)^2=9-(x-3)^2=27-(x+3)^2[/MATH]
This simplifies to:

[MATH](x+3)^2-(x-3)^2=18[/MATH]
[MATH](2x)(6)=18[/MATH]
[MATH]x=\frac{3}{2}[/MATH]
Can you now find $y$?

 

[iuhn]

It took me a while, but yes I can find y now! Knowing what x is, I plugged it into one of the previous equations (this was my process, and please excuse me for not knowing how to type it out as clearly as you typed out yours):

(1.5 - 3)^2 + (y + 1)^2 = 9

2.25 + (y + 1)^2 = 9

(y + 1)^2 = 6.75

y + 1 = 2.60

y = 1.60

Thank you so much, MarkFL! Actually made my day learning exactly how to do this darned problem haha

 

[MarkFL]

There are actually two values of $y$ that would work. Let;s back up to:

[MATH](y+1)^2=\frac{27}{4}[/MATH]
Now, taking the square root of both sides, we should write

[MATH]y+1=\pm\frac{3\sqrt{3}}{2}[/MATH]
Hence:

[MATH]y=-1\pm\frac{3\sqrt{3}}{2}[/MATH]
Now, from the diagram, it appears they want the solution where $-1<y$, but that's not actually stated, and we should be aware that there are two solutions to the problem: