If the angle of elevation of a tower from a distance of 100 metres from its foot is $60^\circ$, then the height of the tower is

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KollegeApply · Accepted Answer

Let AB be the height of tower is h meters Given that: angle of elevation is $60^\circ$ from tower of foot and distance BC=100 meters. Here, we have to find the height of tower. So we use trigonometric ratios. In a triangle ABC $\Rightarrow$ tan C= (AB)/(BC) $\Rightarrow$tan $60^\circ$= (AB)/(BC) $\Rightarrow$ sqrt3= h/100 $\Rightarrow$ h= 100sqrt3 [Solution diagram: 63472_solution_1.png]

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