A 1.5 m tall boy is looking at the top of a temple which is 30 meter in height from a point at certain distance .The angle of elevation from his eye to
the top of the crown of the temple increases from 30° to 60° as he walks towards the temple. Find the distance he walked towards the temple.

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2014-12-19T23:58:32+05:30

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Please see the diagram enclosed.

Boy is at AB.  The temple is PQ.  Angle of elevation changes from 30 to 60 as boy's head (eyes) move from B to S.  We have to find  distance x. (=  BS)

   tan B = tan 30⁰ = 1/√3 =  RQ / RB = 30 / D
             => D = 30 √3 meters

   tan S = tan 60° = √3  =  RQ / RS  =  30 / (30√3 - x)
             30*√3*√3 - x√3 =  30

           x = 60/√3 = 20√3 meters.


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2014-12-20T09:24:22+05:30
Height of boy = 1.5m
Height of temple from ground 30m
Height of temple from boys vision= 30-1.5=28.5 m
Let distance between boy and temple be x.
Using trigonometry we have tan 30°=28.5/x                                                                                                        ⇒  1/√3 = 28.5/x                                                                                                         ⇒    x = 49.26             
When the angle of elevation is
60° boy has walked distance =y                                                                    we have tan60 = 28.5/(49.26-y)                                                                                             ⇒     √3 = 28.5/(49.26-y)                                                                                         ⇒  49.26-y=28.5/√3=16.45                                                                                                        y= 32.81
He has walked 32.81m toward temple. 
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