A pole leans away from the sun at an angle of 5 degrees to the vertical. When the elevation of the sun is 56 degrees, the pole casts a shadow 43 ft long on level ground. How long is the pole?

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Answer 1 · 2018-03-14T15:05:12

Length of the pole #~~56.64ft#

drawn

In figure #AD# is the length of the perpendicular drawn from hugest point A of the leaning pole AB pole to the ground.

#BC=43ft# is the length of the shadow of the leaning pole when angle elevation of the sun is #56^@#

So #AD=ABcos5^@ and BD= AB sin5^@#

Now #(AD)/(DC)=tan56^@#

#=>(ABcos5^@)/(DC)=tan56^@#

#=>(ABcos5^@)/(BC-BD)=tan56^@#

#=>(ABcos5^@)/(43-ABsin5^@)=tan56^@#

#=>(43-ABsin5^@)/(ABcos5^@)=cot56^@#

#=>43/(ABcos5^@)-tan5^@=cot56^@#

#=>43/(ABcos5^@)=cot56^@+tan5^@#

#=>(ABcos5^@)/43=1/(cot56^@+tan5^@)#

#=>ABcos5^@=43/(cot56^@+tan5^@)#

#=>AB=43/(cos5^@(cot56^@+tan5^@))#

#=>AB=43/(cos5^@cot56^@+sin5^@)~~56.64ft#

Alternative way

#(AB)/(sin56^@)=43/(sinangleBAC)=43/sin39^@#

#=>AB=43/(sin39^@)xxsin56^@~~54.64ft#

A pole leans away from the sun at an angle of 5 degrees to the vertical. When the elevation of the sun is 56 degrees​, the pole casts a shadow 43 ft long on level ground. How long is the​ pole?