(B) Let $h$ be the height of the house and $H$ be the height of the tower above the top of the house.From the triangle formed by the angle of depressi

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(B) Let $h$ be the height of the house and $H$ be the height of the tower above the top of the house.From the triangle formed by the angle of depression,we have $\tan(30^\circ) = \frac{h}{12}$.Thus,$h = 12 \tan(30^\circ) = 12 \times \frac{1}{\sqrt{3}} = 4\sqrt{3} \ m$.From the triangle formed by the angle of elevation,we have $\tan(60^\circ) = \frac{H}{12}$.Thus,$H = 12 \tan(60^\circ) = 12 \sqrt{3} \ m$.The total height of the tower is $h + H = 4\sqrt{3} + 12\sqrt{3} = 16\sqrt{3} \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Medium

The angle of elevation of the top of a tower from the top of a house is $60^\circ$ and the angle of depression of its base is $30^\circ$. If the horizontal distance between the house and the tower is $12 \ m$,then the height of the tower is

A
$48 \ \sqrt{3} \ m$
B
$16 \ \sqrt{3} \ m$
C
$24 \ \sqrt{3} \ m$
D
$16 / \sqrt{3} \ m$

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Trigonometrical Equations

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Height and Distance

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The angle of elevation of the top of a tower from the top of a house is $60^\circ$ and the angle of depression of its base is $30^\circ$. If the horizontal distance between the house and the tower is $12 \ m$,then the height of the tower is

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$A$ person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $60^\circ$. When he moves $40 \ m$ away from the bank,he finds the angle to be $30^\circ$. The breadth of the river is.....$m$

From a $60 \ m$ high tower,the angles of depression of the top and bottom of a house are $\alpha$ and $\beta$ respectively. If the height of the house is $\frac{60 \sin(\beta - \alpha)}{x}$,then $x =$

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