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If the elevation of the sun changed from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, made at these two positions is:

  1. 7.5 m

  2. 15 m

  3. 535\sqrt{3} m

  4. 10310\sqrt{3} m

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Answer

If the elevation of the sun changed from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high, made at these two positions is: Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

Height of pole AB = 15 m

Let D be the position where angle of elevation is 30° and C be the point where angle of elevation is 60°.

In triangle ABC,

We know that,

tan60=PerpendicularBase=ABBC3=15BCBC=153BC=153×33BC=1533BC=53 m.\Rightarrow \tan 60^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \dfrac{15}{BC} \\[1em] \Rightarrow BC = \dfrac{15}{\sqrt3} \\[1em] \Rightarrow BC = \dfrac{15}{\sqrt{3}} \times \dfrac{\sqrt3}{\sqrt3} \\[1em] \Rightarrow BC = \dfrac{15\sqrt3}{3} \\[1em] \Rightarrow BC = 5\sqrt3 \text{ m.}

In triangle ABD,

We know that,

tan30=PerpendiuclarBase=ABBD13=15BDBD=153 m.\Rightarrow \tan 30^{\circ} = \dfrac{\text{Perpendiuclar}}{\text{Base}} = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{15}{BD} \\[1em] \Rightarrow BD = 15\sqrt3 \text{ m.}

From figure,

CD = BD - BC

CD = 1535315\sqrt3 - 5\sqrt3

CD = 103 m.10\sqrt3 \text{ m.}

Hence, option 4 is the correct option.

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