# Thread: Equilateral triangle and perpendicular line

1. ## Equilateral triangle and perpendicular line

I didn't quite know what to put as title, so I'm sorry if the title doesn't give a clue of what is to be found

Well, here's the question:

Qu: The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR so that ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, the length of ST is

A. $\frac{1}{2}$

B. $2-\sqrt{3}$

C. $2\sqrt{3} - 3$

D. $2(2-\sqrt{3})$

E. $\frac{1}{3} \sqrt{3}$

I proceeded with finding the angles, angle STR being 150 degrees and angle TSR = angle TRS = 15 degrees.

Then, I found the length of QS, since QRS is a right angle with one of the angles being 45 degrees. QS = $\frac{1}{\sqrt{2}}$ = SR.

Then, fron the sine rule, I know that:
$\frac{L}{sin(15)} = \frac{1}{\sqrt{2} sin(150)}$

So,

$L = \frac{2}{\sqrt{2}sin(15)}$

Now, I don't know the exact form on sin(15) and I don't see any square root of 2 in any of the choices. Is there a quicker/simpler way to solve this?

Thanks!

2. Originally Posted by Unknown008
Qu: The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR so that ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, the length of ST is
Unless I made some error, the answer is just 1/2, because consider that triangle SQT must also be equilateral, therefore length of QT = length of ST = length of TR.

3. No, I don't think that triangle SQT is equilateral, because ST is perpendicular to PQ. Since one of the angles is 90 degrees, the triangle cannot be equilateral.

4. Originally Posted by Unknown008
No, I don't think that triangle SQT is equilateral, because ST is perpendicular to PQ. Since one of the angles is 90 degrees, the triangle cannot be equilateral.
Oh yes I made the error of reading "ST perpendicular to PQ" as "ST parallel to PR". My mistake.

5. So, triangle SQT is a 30-60-90 triangle, and if we label the hypotenuse QT as having length y and TR as having length x, we get cos(pi/6) = sqrt(3)/2 = x/y which gives y = 2x/sqrt(3). So x+2x/sqrt(3) = 1. After some algebraic manipulation I get answer C.

6. Maybe it was hard to read?

EDIT: Sorry, I switched Q and R, so my letters are all messed up... uh, let me see what I should have done.

$\cos(30^\circ)=\frac{\sqrt{3}}{2}=\frac{x}{y}$

$y=\frac{2}{\sqrt{3}}x$

$x+y=x+\frac{2}{\sqrt{3}}x=1$

$x=\dfrac{1}{1+\frac{2}{\sqrt{3}}}$

$=\dfrac{1}{\frac{\sqrt{3}+2}{\sqrt{3}}}$

$=\frac{\sqrt{3}}{\sqrt{3}+2}$

$=\frac{(\sqrt{3})(2-\sqrt{3})}{2^2-3}$

$=2\sqrt{3}-3$

Why would triangle SQT be 30-60-90? It could very well be any other angle

[sorry the 'Q' disappeared]

EDIT: Ok, the triangle QRT is 30-60-90, so, STR is also 30-60-90.

8. Uh... undefined, T is on PR, not QR.

9. Originally Posted by Unknown008
Uh... undefined, T is on PR, not QR.
Yes, I edited my post, and many apologies for my not being able to read.

But I think the calculations work out the same, by coincidence. Have I made yet another stupid mistake?

10. Shoo! My drawing is not good. It's not TQ which is equal to ST but TR... darn...

11. Finally, I hope I get it right. Yes, now I understand your method.

So,

$y + x = 1$

$\frac{2x}{\sqrt{3}} + x = 1$

$x = \frac{1}{\frac{2}{\sqrt{3}} + 1}$

$x = 2\sqrt{3} - 3$

Phew, thanks!

12. Hello,
Solvable by trig yes but I say geometry no. A quick scaled triangle shows ST = approx 1/2 of QR.

bjh

13. You're right, but all of C, D and E answers are approximately 0.5. How would you choose from those three?

14. Hello unknown008,
Solve using trig to get the answer

bjh

15. Well, post what you would have done using trigonometry, I'll be happy to see how you would do it.

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