A person of height 6 ft wants to pluck a fruit which is on \[\frac{26}{3}\text{ft}\] a high tree. If the person is standing \[\frac{8}{\sqrt{3}}\text{ft}\] away from the base of the tree, then at what angle should be throw, so that it hits the fruit? [SSC (CGL) Pre 2015] |
A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) \[75{}^\circ \]
Correct Answer: A
Solution :
Let the required angle be\[\theta .\] |
\[DC=EB=\frac{8}{\sqrt{3}}\text{ft}\] |
Here, \[AB=AC-BC=\frac{26}{3}-6=\frac{26-18}{3}=\frac{8}{3}\text{ft}\] |
In \[\Delta ABE,\]\[tan\theta =\frac{AB}{BE}\,\,=\,\,\frac{\frac{8}{3}}{\frac{8}{\sqrt{3}}}\,\,=\,\,\frac{1}{\sqrt{3}}\] |
\[\Rightarrow \] \[\tan \theta =\frac{1}{\sqrt{3}}\] |
\[\therefore \] \[\theta =30{}^\circ \] |
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