A) \[50\sqrt{3}\text{ }meters\]
B) \[\frac{50}{\sqrt{3}}\text{ }meters\]
C) \[100\sqrt{3}\text{ }meters\]
D) \[\frac{100}{\sqrt{3}}\text{ }meters\]
Correct Answer: A
Solution :
Let \[AB=h\] be the height of the tower, and let \[AD=x\text{ }m\] From right angled \[\Delta \,ABD,\] \[\tan {{60}^{o}}=\frac{AB}{AD}\] or \[\sqrt{3}=\frac{h}{x}\] or \[x=\frac{h}{\sqrt{3}}\] ?..(i) From right angled \[\Delta \,BCA,\] \[\tan {{30}^{o}}=\frac{AB}{CA}\] or \[\frac{1}{\sqrt{3}}=\frac{h}{100+x}\] or \[h=\frac{100+x}{\sqrt{3}}=\frac{100+\frac{h}{\sqrt{3}}}{\sqrt{3}}\] or \[\sqrt{3}\,h=100+\frac{h}{\sqrt{3}}\] or \[\frac{2}{\sqrt{3}}h=100\] \[\therefore \] \[h=50\sqrt{3}\,m\]You need to login to perform this action.
You will be redirected in
3 sec