A tower subtends an angle of $30^\circ$ at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is $60^\circ$. The height of the tower is

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KollegeApply · Accepted Answer

Let AB be the tower and C is a point on the same level as its foot such that $\angle$ACB = $30^\circ$ The given situation can be represented as, Here D is a point h m above the point C. In ΔBCD, $\Rightarrow$ tan B=(CD)/(CB) $\Rightarrow$ tan $60^\circ$=h/(CB) $\Rightarrow$ sqrt3=h/(CB) $\Rightarrow$ CB=h/sqrt3 Again in triangle ABC, tan C=(AB)/(CB) $\Rightarrow$ tan30= (AB)/((h/sqrt3)) [Using (1)] $\Rightarrow$1/sqrt3=AB/((h/sqrt3)) $\Rightarrow$ AB=h/3 [Solution diagram: 63559_solution_1.png]

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