The Question
The diagram shows a square P Q R S PQRS PQRS T T T S R SR SR U U U R Q RQ RQ ∠ S P T \angle SPT ∠ SPT ∠ T P U \angle TPU ∠ TP U
What is the length of R U RU R U
Identities to Know
tan ( 90 − θ ) = cot ( θ ) cot ( θ ) = 1 tan ( θ ) tan ( 2 θ ) = 2 tan ( θ ) 1 − tan 2 ( θ ) \begin{align*}
\tan(90 - \theta) &= {\cot(\theta)}\\
\cot(\theta) &= \frac{1}{\tan(\theta)}\\
\tan(2\theta) &= \frac{2\tan(\theta)}{1-\tan^2(\theta)}\\
\end{align*} tan ( 90 − θ ) cot ( θ ) tan ( 2 θ ) = cot ( θ ) = tan ( θ ) 1 = 1 − tan 2 ( θ ) 2 tan ( θ ) Solution with Trigonometry
Let α = ∠ S P T , β = ∠ U P Q , Q U = y , U R = x \alpha = \angle SPT, \beta = \angle UPQ, QU = y, UR = x α = ∠ SPT , β = ∠ U PQ , Q U = y , U R = x
tan ( α ) = S T S P = 1 2 ∴ α = tan − 1 ( 1 2 ) \begin{align*}
\tan(\alpha) &= \frac{ST}{SP} = \frac{1}{2}\\
\therefore\space\alpha &= \tan^{-1}(\frac{1}{2})
\end{align*}\\ tan ( α ) ∴ α = SP ST = 2 1 = tan − 1 ( 2 1 ) Now, because tan ( β ) = U Q Q P = y 2 \tan(\beta) = \frac{UQ}{QP} = \frac{y}{2} tan ( β ) = QP U Q = 2 y β \beta β y y y
β = 90 − 2 α tan ( β ) = tan ( 90 − 2 α ) = cot ( 2 α ) = 1 tan ( 2 α ) L e t t = tan ( 2 α ) = 2 tan ( α ) 1 − tan 2 ( α ) ∵ α = tan − 1 ( 1 2 ) t = 2 ⋅ tan ( tan − 1 ( 1 2 ) ) 1 − [ tan ( tan − 1 ( 1 2 ) ) ] 2 \begin{align*}
\beta &= 90 - 2\alpha\\ \\
\tan(\beta)&=\tan(90 - 2\alpha)=\cot(2\alpha)\\
&=\frac{1}{\tan(2\alpha)}\\ \\
Let\space t &= \tan(2\alpha)\\
&=\frac{2\tan(\alpha)}{1 - \tan^2(\alpha)}\\
\because\space\alpha &= \tan^{-1}(\frac{1}{2})\\
t&=\frac{2 \cdot \tan(\tan^{-1}(\frac{1}{2}))}{1 - [\tan(\tan^{-1}(\frac{1}{2}))]^{2}}
\end{align*} β tan ( β ) L e t t ∵ α t = 90 − 2 α = tan ( 90 − 2 α ) = cot ( 2 α ) = tan ( 2 α ) 1 = tan ( 2 α ) = 1 − tan 2 ( α ) 2 tan ( α ) = tan − 1 ( 2 1 ) = 1 − [ tan ( tan − 1 ( 2 1 )) ] 2 2 ⋅ tan ( tan − 1 ( 2 1 )) Notice that tan \tan tan tan − 1 \tan^{-1} tan − 1
∴ t = 2 ⋅ 1 2 1 − ( 1 2 ) 2 = 1 ( 3 4 ) t = 4 3 \begin{align*}
\therefore\space t&=\frac{2 \cdot \frac{1}{2}}{1 - (\frac{1}{2})^2}\\
&=\frac{1}{(\frac{3}{4})}\\
t&=\frac{4}{3}
\end{align*} ∴ t t = 1 − ( 2 1 ) 2 2 ⋅ 2 1 = ( 4 3 ) 1 = 3 4 Now, going back to finding tan ( β ) \tan(\beta) tan ( β )
tan ( β ) = 1 t = 3 4 \begin{align*}
\tan(\beta) &= \frac{1}{t}\\
&=\frac{3}{4}
\end{align*} tan ( β ) = t 1 = 4 3 That means:
3 4 = U Q Q P = y 2 ∴ y = 3 2 ∵ x = 2 − y x = 1 2 ⟺ R U = 1 2 \begin{align*}
\frac{3}{4}&=\frac{UQ}{QP}=\frac{y}{2}\\
\therefore y &= \frac{3}{2}\\ \\
\because x &= 2 - y\\
x &= \frac{1}{2} \Longleftrightarrow RU = \frac{1}{2}
\end{align*} 4 3 ∴ y ∵ x x = QP U Q = 2 y = 2 3 = 2 − y = 2 1 ⟺ R U = 2 1 Conclusion
There are many ways to solve a single geometry problem, thus you are encouraged to find different solutions to a question, and this is very helpful for enhancing your problem solving skills.
