Questions involving right-angle triangles can often be solved by applying Pythagorean thm or special-angle rules.
We'll look for these rules, a Precise approach.

Since \(\angle\)QPT = 30, then \(\angle\)QST=60 and we can also fill in \(\angle\)RTS=30 and \(\angle\)RQT =30. That is, all our triangles are 30-60-90 triangles.
This means that the ratio between the short leg and the hypotenuse is 1:2 and the long leg to the short leg is sqrt(3):1.
So:
PQ = 1 --> QT = 1/2
QT = 1/2 --> RT = 1/4
RT = 1/4 --> RS = 1/(4*sqrt(3)) = sqrt(3)/sqrt(3) * 1/(4*sqrt(3)) = sqrt(3)/12.

(B) is our answer.
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I think so. Try to rationalize RS = 1/(4*sqrt(3)).
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Ah, didn't think to do that. Thanks
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Option B using 30-60-90

can we use the formula:- QS * RS=TS^2 ?

Calculations here are quite simple because all the triangles are 30-60-90
Sides opposite those angles are in ratio \(x : x\sqrt{3} : 2x\)

We are given that the angle at vertex P = 30°.
The angle at P starts a chain in which we find nothing except 30-60-90 angle possibilities
∠PTQ = 90°. The third angle must = 60° (∠PQT)
In turn, that 60° is part of a 90° angle, so adjacent ∠SQT = 30°
In turn, ∆QRT 's second angle = 90° . . . etc.

Halve two sides (PQ and QT), divide the third (RT) by \(\sqrt{3}\),
and we have RS.

(1) For ∆ PQS, side \(PQ = 1\)
PQ, opposite the 90° angle, corresponds with \(2x\) in the ratio of sides
\(2x = 1\)
\(x = \frac{1}{2}\)
\(x\) corresponds with QT, the side opposite the 30° angle.
\(QT = \frac{1}{2}\)

(2) For ∆ QST, side \(QT = \frac{1}{2}\)
QT, opposite the 90° angle, corresponds with \(2x\)
RT, opposite the 30° angle, corresponds with \(x\)
So \(RT = \frac{1}{2}QT\)
\(QT = \frac{1}{2}\)
\(RT =( \frac{1}{2}* \frac{1}{2}) = \frac{1}{4}\)

(3) For ∆, side \(RT = \frac{1}{4}\)
RT, opposite the 60° angle, corresponds with \(x\sqrt{3}\)
RT = \(\frac{1}{4} = x\sqrt{3}\)

\(RS = x = \frac{1}{4\sqrt{3}}= RS,\)opposite the 30° angle

\(x = (\frac{1}{4\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}})\)

\(x= RS = \frac{\sqrt{3}}{12}\)

Answer B
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All 3 triangles are 30-60-90.

And they are all similar.

We can find the length of one more side in the larger triangle to determine the similar triangle ratio and one side of the triangle QTR. We can then use the properties of similar triangles to determine the side RS
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