(B) Let the height of the cliff be $h = 500 \ m$.Let the diameter of the base be $D = d_1 + d_2$.From the right-angled triangles formed:$d_1 = h \cot(

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(B) Let the height of the cliff be $h = 500 \ m$.Let the diameter of the base be $D = d_1 + d_2$.From the right-angled triangles formed:$d_1 = h \cot(60^\circ) = 500 \times \frac{1}{\sqrt{3}} = \frac{500}{\sqrt{3}} \ m$.$d_2 = h \cot(30^\circ) = 500 \times \sqrt{3} = 500\sqrt{3} \ m$.The diameter $D = d_1 + d_2 = \frac{500}{\sqrt{3}} + 500\sqrt{3}$.$D = \frac{500 + 500(3)}{\sqrt{3}} = \frac{500 + 1500}{\sqrt{3}} = \frac{2000}{\sqrt{3}} \ m$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Medium

The base of a cliff is circular. From the extremities of a diameter of the base,the angles of elevation of the top of the cliff are $30^\circ$ and $60^\circ$. If the height of the cliff is $500 \ m$,then the diameter of the base of the cliff is:

A
$1000 \sqrt{3} \ m$
B
$2000 / \sqrt{3} \ m$
C
$1000 / \sqrt{3} \ m$
D
$2000 \sqrt{2} \ m$

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The base of a cliff is circular. From the extremities of a diameter of the base,the angles of elevation of the top of the cliff are $30^\circ$ and $60^\circ$. If the height of the cliff is $500 \ m$,then the diameter of the base of the cliff is:

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The angle of elevation of a stationary cloud from a point $2500 \, m$ above a lake is $15^\circ$ and the angle of depression of its reflection in the lake is $45^\circ$. The height of the cloud above the lake level is

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