A) 10 m
B) 15 m
C) \[15\sqrt{3}\,\,m\]
D) 30 m
Correct Answer: C
Solution :
Let AB is tower, the angles of elevation of the top of a vertical tower AB from two points D and C are \[{{30}^{o}}\] and \[{{60}^{o}}\] respectively. \[CD=30\]metre Let the height of tower be h metres and \[BC=x\text{ }m\] In triangle \[ABC,\frac{AB}{BC}=\tan {{60}^{o}}\] \[\frac{h}{x}=\sqrt{3}\] ?..(i) In triangle \[ABD,\frac{AB}{BD}=\tan {{30}^{o}}\] \[\therefore \] \[\frac{h}{(x+30)}=\frac{1}{\sqrt{3}}\] or \[\sqrt{3h}=x+30\] Put value of x from equation (i) in equation (ii), we get \[\sqrt{3h}=\frac{h}{\sqrt{3}}+30\] or \[\sqrt{3h}-\frac{h}{\sqrt{3}}=30\] or \[\frac{2h}{\sqrt{3}}=30\] \[\therefore \] \[h=\frac{30\sqrt{3}}{2}=15\sqrt{3}\] metres So, height of the tower is \[15\sqrt{3}\] metre.You need to login to perform this action.
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