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The angle of elevation of the top of a tower as observed from a point in a horizontal plane through the foot of the tower is $30^\circ$. When the observer moves towards the tower a distance of $100$ m, he finds the angle of elevation of the top to be $60^\circ$. Find the height of the tower and the distance of first position from the tower.
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Solution:
(a)
In $\triangle DAB$,
$\frac{h}{x} = \tan 60^\circ$
$h = \sqrt{3}x$ ............(i) (1 Mark)
In $\triangle CAB$,
$\frac{h}{x+100} = \tan 30^\circ$
$\sqrt{3} h = x + 100$ ...........(ii) (1.5 Mark)
Solving (i) and (ii), we get
$x = 50 \Rightarrow AC = 150$ m (1 Mark)
$h = 50\sqrt{3}$ m or $86.5$m (1.5 Mark)
(a)
In $\triangle DAB$,
$\frac{h}{x} = \tan 60^\circ$
$h = \sqrt{3}x$ ............(i) (1 Mark)
In $\triangle CAB$,
$\frac{h}{x+100} = \tan 30^\circ$
$\sqrt{3} h = x + 100$ ...........(ii) (1.5 Mark)
Solving (i) and (ii), we get
$x = 50 \Rightarrow AC = 150$ m (1 Mark)
$h = 50\sqrt{3}$ m or $86.5$m (1.5 Mark)