Question

the angle of depression from the top of a vertical tower to a point on the ground is found to be 60°and from a point 50m above the foot of the tower angle of depression to the same point to be 30° find the height of the tower​

Answer 1

Height of the tower, H = (h+10)m

In △ABC,

tan60° =  $\frac{(h+10)}{AB}$

$AB = \frac{(h+10)}{\sqrt{3}} ...(i)$

In △DCO,

tan45° = $\frac{h}{ab}$

∴ AB = h⋯(ii)

comparing eq.(i) and (ii), we get :

h = $\frac{(h+10)}{\sqrt{3} }$

∴ $\sqrt{3}h$ = h+10

⇒ $\sqrt{3}h$ - h = 10

h $\sqrt{3-1} = 10$

∴ h = $\frac{10}{\sqrt{3-1} }$

Height of tower = h + 10

= $\frac{10}{\sqrt{3-1} }$ +10

= $\frac{10+10\sqrt{3}-10}{\sqrt{3}-1 }$

H = $\frac{10 \sqrt{3}}{\sqrt{3} -1}$m

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