Answer:
Let AB and CD be two towers of height x and y respectively. M is the mid-point of BC i.e., \[BM=MC\] In \[\Delta \text{ }ABM\], we have \[\frac{AB}{BM}=\tan \,30{}^\circ \] \[BM=\frac{x}{\tan \,30{}^\circ }\] ?(i) In \[\Delta \text{ }CDM\], we have \[\frac{DC}{MC}=\tan \,\,60{}^\circ \] \[\frac{y}{MC}=\tan \,\,60{}^\circ \] \[MC=\frac{y}{\tan \,\,60{}^\circ }\] ?(ii) From eq. (i) and (ii), we get \[\frac{x}{\tan 30{}^\circ }=\frac{y}{\tan 60{}^\circ }\] \[\frac{x}{y}=\frac{\tan 30{}^\circ }{\tan 60{}^\circ }\] \[\frac{x}{y}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}\] \[\therefore x:y=1:3\]
You need to login to perform this action.
You will be redirected in
3 sec