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Question
In Fig. 1, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the length of the ladder. (use3√=1.73)
Solution
In the given figure,
AB = AD + DB = 6 m
Given: AD = 2.54 m
⇒ 2.54 m + DB = 6 m
⇒ DB = 3.46 m
Now, in the right triangle BCD,
`(BD)/(CD)=sin 60^@`
`=>(3.46m)/(CD)=sqrt3/2`
`=>(3.46m)/(CD)=1.73/2`
`=>CD=(2xx3.46m)/1.73`
⇒CD=4 m
Thus, the length of the ladder CD is 4 m.
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