question_answer

The angle of depressions of the top and bottom of 10 m tall building from the top of a multistoried building are \[{{30}^{o}}\] and \[{{60}^{o}}\] respectively. Find the height of the multistoried building and the distance between the two buildings.

A)  \[15\,m,\,\,5\sqrt{3}\,m\]  

B)         \[15\,m,\,\,6\sqrt{3}\,m\]

C)         \[16\,m,\,\,4\sqrt{3}\,m\]  

D)         \[16\,m,\,\,5\sqrt{3}\,m\]

Correct Answer: A

Solution :

Let AB be the multi storied building of height h and let the distance between two building be x metres. \[\angle XAC=\angle ACB={{60}^{o}}\](Alternate angles) \[\angle XAD=\angle \text{ }ADE={{30}^{o}}\] (Alternate angles) In \[\Delta ADE,\] \[\tan {{30}^{o}}=\frac{AE}{ED}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h-10}{x}\]                         \[[\,\,\,CB=DE=x]\] \[\Rightarrow \]            \[x=\sqrt{3}\,(h-10)\]        ?..(1) In  \[\Delta \,ACB,\] \[\tan {{60}^{o}}=\frac{h}{x}\Rightarrow \sqrt{3}=\frac{h}{x}\Rightarrow x=\frac{h}{\sqrt{3}}\]        ..(2) From (1) and (2), we have \[\sqrt{3}\,(h-10)=\frac{h}{\sqrt{3}}\Rightarrow h=15\,m\] From (2), \[x=\frac{h}{\sqrt{3}}\]So, \[x=\frac{15}{\sqrt{3}}=5\sqrt{3}\,m\] Hence, height of multi storied building = 15 metres Distance between two buildings \[=5\sqrt{3}\] metres.