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
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