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A straight road leads to foot of a tower whose shadow is found to be $40$ m longer when sun's altitude is $30^\circ$ than when it is $60^\circ$. Find the height of the tower and length of shadow in both the situations. (Use $\sqrt{3} = 1.73$)
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Solution: For correct figure: (1 Mark)
Let AB be the tower
In $\triangle BAC$, $\tan 60^\circ = \frac{h}{x}$ (1/2 Mark)
$\Rightarrow h = x\sqrt{3}$ ..... (i) (1/2 Mark)
In $\triangle BAD$, $\tan 30^\circ = \frac{h}{40+x}$ (1/2 Mark)
$\Rightarrow 40 + x = h\sqrt{3}$ ..... (ii) (1/2 Mark)
Using (i) and (ii) $x = 20$ m (1 Mark)
... Length of the shadow at $60^\circ$ is $20$ m and at $30^\circ$ is $60$ m (1/2 Mark)
and $h = 20\sqrt{3} = 34.6$ m. (1/2 Mark)
Let AB be the tower
In $\triangle BAC$, $\tan 60^\circ = \frac{h}{x}$ (1/2 Mark)
$\Rightarrow h = x\sqrt{3}$ ..... (i) (1/2 Mark)
In $\triangle BAD$, $\tan 30^\circ = \frac{h}{40+x}$ (1/2 Mark)
$\Rightarrow 40 + x = h\sqrt{3}$ ..... (ii) (1/2 Mark)
Using (i) and (ii) $x = 20$ m (1 Mark)
... Length of the shadow at $60^\circ$ is $20$ m and at $30^\circ$ is $60$ m (1/2 Mark)
and $h = 20\sqrt{3} = 34.6$ m. (1/2 Mark)