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 \[4\,m\].
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