The angle of elevation of a stationary cloud from a point $25 m$ above a lake is $15^{o}$ and the angle of depression of its image in the lake is $45^{o}$ . The height of the cloud above the lake level is

25 m

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25 \sqrt{3} m

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50 m

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50 \sqrt{3} m

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Solution

The correct option is B $25 \sqrt{3} m$
$tan 15 ° = 2 - \sqrt{3} = \frac{x}{x + 50} \Rightarrow x = \frac{- 50 (2 - \sqrt{3})}{1 - \sqrt{3}} N o w, x + 25 \Rightarrow \frac{- 50 (2 - \sqrt{3})}{1 - \sqrt{3}} + 25 = 25 \sqrt{3}$

Therefore, Height of Cloud is $25 \sqrt{3}$

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Similar questions

Q. The angle of elevation of a stationary cloud from a point

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and the angle of depression of its reflection in the lake is

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Q. The angle of elevation of stationary cloud from a point 25 ml above the lake is

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The angle of elevation of a stationery cloud from a point 2500 m above a lake is $15^{\circ}$ and the angle of depression of its reflection in the lake is $45^{\circ}$ . What is the height of the cloud above the lake level? (Use tan $15^{\circ}$ = 0.268)

The angle of elevation of a cloud from a height h above the level of water in a lake is $α$ and the angle of depression of its image in the lake is $β$ . Find the height of the cloud above the surface of the lake :

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The angle of elevation of a stationary cloud from a point 25m above a lake is 15o and the angle of depression of its image in the lake is 45o. The height of the cloud above the lake level is

The correct option is B 25√3 mtan15°=2−√3=xx+50⇒x=−50(2−√3)1−√3Now, x+25⇒−50(2−√3)1−√3+25=25√3Therefore, Height of Cloud is 25√3

The angle of elevation of a stationary cloud from a point $25 m$ above a lake is $15^{o}$ and the angle of depression of its image in the lake is $45^{o}$ . The height of the cloud above the lake level is

The correct option is B $25 \sqrt{3} m$
$tan 15 ° = 2 - \sqrt{3} = \frac{x}{x + 50} \Rightarrow x = \frac{- 50 (2 - \sqrt{3})}{1 - \sqrt{3}} N o w, x + 25 \Rightarrow \frac{- 50 (2 - \sqrt{3})}{1 - \sqrt{3}} + 25 = 25 \sqrt{3}$

Therefore, Height of Cloud is $25 \sqrt{3}$