Question

A 6 mts. pole casts a shadow of 3.6 mts. at a certain time of the day. Find the length of the shadow cast by a 4.5 mts. tower at the same time.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{two}\mathrm{cases},\mathrm{we}\mathrm{have}, \\ \angle \mathrm{ABC}=\angle \mathrm{PQR}=90° \\ \angle \mathrm{ACB}=\angle \mathrm{PRQ}(\mathrm{Angle}\mathrm{of}\mathrm{elevation}\mathrm{of}\mathrm{the}\mathrm{Sun}\mathrm{at}\mathrm{the}\mathrm{same}\mathrm{time}) \\ \therefore △\mathrm{ABC}~△\mathrm{PQR}(\mathrm{AA}\mathrm{similarity}\mathrm{test}) \\ \mathrm{Then},\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}},\mathrm{where}\mathrm{QR}\mathrm{is}\mathrm{the}\mathrm{shadow}\mathrm{cast}\mathrm{by}\mathrm{the}4.5\mathrm{m}\mathrm{tall}\mathrm{tower} \\ \Rightarrow \frac{6}{4.5}=\frac{3.6}{\mathrm{QR}} \\ \Rightarrow \frac{4.5\times 3.6}{6}=\mathrm{QR}=2.7\mathrm{m} \end{gathered}$$