Question

In fig 8.26, $\angle$C = 90$°$ and D is the mid point of seg AC. Write

(1) $\frac{\tan \angle CAB}{\tan \angle CDB}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \tan \angle \mathrm{CAB}=\frac{\mathrm{opposite}\mathrm{side}\mathrm{of}\angle \mathrm{CAB}}{\mathrm{adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{CAB}}=\frac{\mathrm{BC}}{\mathrm{CA}} \\ \tan \angle \mathrm{CDB}=\frac{\mathrm{opposite}\mathrm{side}\mathrm{of}\angle \mathrm{CDB}}{\mathrm{adjacent}\mathrm{side}\mathrm{of}\angle \mathrm{CDB}}=\frac{\mathrm{BC}}{\mathrm{CD}} \end{gathered}$$

$$\begin{gathered} \mathrm{Consider}\frac{\tan \angle \mathrm{CAB}}{\tan \angle \mathrm{CDB}}=\frac{\mathrm{BC}}{\mathrm{CA}}÷\frac{\mathrm{BC}}{\mathrm{CD}} \\ =\frac{\mathrm{BC}}{\mathrm{CA}}\times \frac{\mathrm{CD}}{\mathrm{BC}} \\ =\frac{\mathrm{CD}}{\mathrm{CA}}=\frac{\mathrm{CD}}{2\mathrm{CD}}=\frac{1}{2}(\because \mathrm{D}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{CA}) \end{gathered}$$