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A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane the angles of elevation of the bottom and top of the flagstaff are ? and 2? respectively. What is the height of the tower ?
Given:A flagstaff of height h is surmounted on a verticle tower.Angles of elevation of the bottom and top of the flagstaff are ? and 2? respectively.Formula used:1. tan ? = Perpendicular/Base2. tan 2? = \(\frac{(2tan ?)}{(1 - tan^2?)} \)3. cos 2? = \( \frac{1-tan^2?}{1+tan^2?}\) Calculation:Let the height of the tower be x.In an right-angles triangle ABCtan ? = AC/AB = x/AB? AB = x/tan ? ------(1)In an right-angles triangle ABDtan 2? = AD/AB = (x + h)/AB? \(\frac{(2tan ?)}{(1 - tan^2?)} \) = (x + h)/AB [Using formula (2)]? \(\frac{(2tan ?)}{(1 - tan^2?)} \) = \(\frac{ (x + h)tan\ ?}{ x} \) [From equation (1)]? \(\frac{(2)}{(1 - tan^2?)} \) = \(1+\frac{ h}{ x} \) ? \(\frac{h}{x} = \frac{2}{1-tan^2?}-1\)? \(\frac{h}{x} = \frac{2-(1-tan^2?)}{1-tan^2?}\)? \(\frac{h}{x} = \frac{1+tan^2?}{1-tan^2?}\) ? \(\frac{x}{h} = \frac{1-tan^2?}{1+tan^2?}\) = cos2? [Using formula (3)]? x = hcos 2? ? The height of the tower is h cos 2?
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