Answer:
Let CB be the tower of x m and AC be the flag staff of 5 m. Then, in \[\Delta \text{ }CPB\] \[\tan \,30{}^\circ =\frac{x}{PB}\] \[PB=\frac{x}{\tan \,30{}^\circ }=\sqrt{3}x\] ?(i) In\[\Delta \text{ }APB\] \[\tan \,60{}^\circ =\frac{5+x}{PB}\] \[PB=\frac{5+x}{\sqrt{3}}\] ?(ii) From eq. (i) and (ii) \[\sqrt{3}x=\frac{x+5}{\sqrt{3}}\] \[\Rightarrow x=\frac{x+5}{3}\] \[\Rightarrow 3x-x=5\] \[\Rightarrow 2x=5\] \[\Rightarrow x=5/2=2.5\] \[\therefore \] Height of the tower is \[2.5\text{ m}\].
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