A) \[\frac{50}{\sqrt{3}}m\]
B) \[\frac{40}{\sqrt{3}}m\]
C) \[40\sqrt{3}\,m\]
D) \[50\sqrt{3}\,m\]
Correct Answer: B
Solution :
Height of tower = 40 m Angle of depression \[={{60}^{o}}\] \[\Rightarrow \] \[\angle CAB={{90}^{o}}-{{60}^{o}}={{30}^{o}}\] In \[\Delta \,ABC,\frac{BC}{AB}=\tan {{30}^{o}}\] \[\Rightarrow \] \[\frac{BC}{40}=\frac{1}{\sqrt{3}}\Rightarrow BC=\frac{40}{\sqrt{3}}m\] Hence, the distance between the foot of the tower and ball \[\frac{40}{\sqrt{3}}m.\]You need to login to perform this action.
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