Equilateral triangle and perpendicular line

**Unknown008** · August 4th 2010, 10:34 AM

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!

**undefined** · August 4th 2010, 10:51 AM

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.

**Unknown008** · August 4th 2010, 10:55 AM

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.

**undefined** · August 4th 2010, 10:57 AM

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.

**undefined** · August 4th 2010, 11:08 AM

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.

**undefined** · August 4th 2010, 11:32 AM

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.

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

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

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

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

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

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

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

$=2\sqrt{3}-3$

**Unknown008** · August 4th 2010, 11:40 AM

Sorry, I can't follow your reasoning here.

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.

**Unknown008** · August 4th 2010, 11:44 AM

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

**undefined** · August 4th 2010, 11:48 AM

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?

**Unknown008** · August 4th 2010, 11:51 AM

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

**Unknown008** · August 4th 2010, 11:56 AM

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!

**bjhopper** · August 4th 2010, 12:41 PM

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

bjh

**Unknown008** · August 4th 2010, 11:38 PM

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

**bjhopper** · August 5th 2010, 11:11 AM

Hello unknown008,
Solve using trig to get the answer

bjh

**Unknown008** · August 12th 2010, 05:46 AM

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

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