Question

If Fig. 10.49, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.

Answer 1

In right ∆PQS,

$$\begin{gathered} {\mathrm{PQ}}^2={\mathrm{PS}}^2+{\mathrm{SQ}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow {\mathrm{PQ}}^2={\left(3\right)}^2+{\left(4\right)}^2 \\ \Rightarrow {\mathrm{PQ}}^2=9+16=25 \\ \Rightarrow \mathrm{PQ}=\sqrt{25}=5\mathrm{cm} \end{gathered}$$

In right ∆PQR,

$$\begin{gathered} \tan \theta =\frac{\mathrm{PQ}}{\mathrm{RQ}} \\ \Rightarrow \tan \theta =\frac{5\mathrm{cm}}{9\mathrm{cm}}=\frac{5}{9} \end{gathered}$$

Thus, the value of tanθ is $\frac{5}{9}$.