A person standing on the bank of a river observes that the angle of elevation of the top of a tower on the opposite bank is 60^°. When he moves 30 m…

Let the height of tower BA be h m and the width of river BC be x m \(\displaystyle \tan 60^{\circ}=\sqrt{3}=\frac{\mathrm{h}}{\mathrm{x}}\) \[\Rightarrow \mathrm{h}=\sqrt{3} \mathrm{x}---(\mathrm{i}) \] \(\displaystyle \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{h}}{30+\mathrm{x}}\) \[\Rightarrow h \sqrt{3}=30+x---(i i) \] Solving (i) and (ii), we get \[x=15 \] and \(\displaystyle \mathrm{h}=15 \sqrt{3}=15 \times 1.732=25.98 \mathrm{~m}\) ∴ Height of tower \(\displaystyle =25.98 \mathrm{~m}\) and width of river \(\displaystyle =15 \mathrm{~m}\)