question_answer

In Fig. 1, AB is a 6 m high pole and CD is a ladder inclined at an angle of \[60{}^\circ \] to the horizontal and reaches up to a point D of pole. If \[AD=2.54\text{ }m\], find the length of the ladder. (Use \[\sqrt{3}=1.73\])

Answer:

Given, \[AB=6\text{ }m\] and \[AD=2.54\text{ }m\].

\[\therefore \,\,DB=(6-2.54)\text{ }m=3.46\text{ }m\]

In \[\Delta \text{ }DBC\],

\[\sin \,60{}^\circ =\frac{DB}{DC}\]

\[\frac{\sqrt{3}}{2}=\frac{3.46}{DC}\]

\[\Rightarrow DC=\frac{3.46\times 2}{1.732}=3.995\,m\simeq 4\,m\]

\[\therefore \] The length of the ladder is \[4\,m\].