A) 50 metres
B) 75 metres
C) 100 metres
D) 150 metres
Correct Answer: B
Solution :
Let \[AC=x\text{ }m\] and \[AB=h\text{ }m\] From \[\Delta s\,\,BCA,\] \[\tan {{30}^{o}}=\frac{AB}{AC}\] or \[\frac{1}{\sqrt{3}}=\frac{h}{x}\] or \[h=\frac{x}{\sqrt{3}}\] From \[\Delta \,BDA\,\,\tan {{60}^{o}}=\frac{AB}{AD}\] or \[\sqrt{3}=\frac{h}{x-50}\] or \[h=(x-50)\,\sqrt{3}\] Equating the two values of h, we get \[\frac{x}{\sqrt{3}}=(x-50)\sqrt{3}\] or \[x=(x-50)\,3=3x-150\] or \[2x=150\] \[\therefore \] \[x=75\,m\]You need to login to perform this action.
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