Bunuel wrote:

If PQ = 1, what is the length of RS ?

A. \(\frac{1}{12}\)

B. \(\frac{\sqrt{3}}{12}\)

C. \(\frac{1}{6}\)

D. \(\frac{2}{3 \sqrt{3}}\)

E. \(\frac{2}{\sqrt{12}}\)

Attachment:

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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