Question

In the given figure. Find RP and PS using the information given in ∆PSR.

Answer 1

In ∆PSR,
∠S = 90^∘, ∠P = 30^∘, ∴ ∠R = 60^∘

By 30^∘ − 60^∘ − 90^∘ theorem,

$$\begin{gathered} \mathrm{RS}=\frac{1}{2}\times \mathrm{RP} \\ \Rightarrow 6=\frac{{\displaystyle 1}}{{\displaystyle 2}}\times \mathrm{RP} \\ \Rightarrow 6\times 2=\mathrm{RP} \\ \Rightarrow \mathrm{RP}=12...\left(1\right) \\ \mathrm{PS}=\frac{\sqrt{3}}{2}\times \mathrm{RP} \\ =\frac{\sqrt{3}}{2}\times 12 \\ =6\sqrt{3}...\left(2\right) \end{gathered}$$

Hence, RP = 12 and PS = 6$\sqrt{3}$.