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(B) Let the height of the tower be $h \ m$.Let the length of the shadow when the sun's altitude is $60^{\circ}$ be $x \ m$.In the right-angled triangl

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$20 \sqrt{3}$

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(B) Let the height of the tower be $h \ m$.Let the length of the shadow when the sun's altitude is $60^{\circ}$ be $x \ m$.In the right-angled triangle formed by the tower and the shadow at $60^{\circ}$,we have $	an 60^{\circ} = h / x$,so $x = h / \sqrt{3}$.When the sun's altitude is $30^{\circ}$,the length of the shadow is $(x + 40) \ m$.In the right-angled triangle formed by the tower and the shadow at $30^{\circ}$,we have $	an 30^{\circ} = h / (x + 40)$,so $x + 40 = h / 	an 30^{\circ} = h \sqrt{3}$.Substituting $x = h / \sqrt{3}$ into the equation: $h / \sqrt{3} + 40 = h \sqrt{3}$.$40 = h \sqrt{3} - h / \sqrt{3} = h (\sqrt{3} - 1 / \sqrt{3}) = h ( (3 - 1) / \sqrt{3} ) = 2h / \sqrt{3}$.$2h = 40 \sqrt{3}$,which gives $h = 20 \sqrt{3} \ m$.

The shadow of a tower standing on a level ground is found to be $40 \ m$ longer when the sun's altitude is $30^{\circ}$ than when it is $60^{\circ}$. Find the length of the tower. (In $m$)

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