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A man on the top of a cliff 100 meter hi...

A man on the top of a cliff 100 meter high, observes the angles of depression of two points on the opposite sides of the cliff as `30^(@)` and `60^(@)` respectively. Find the difference between the two points.

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To solve the problem step by step, we will use the concepts of trigonometry, specifically the tangent function, to find the distances from the cliff to the two points observed by the man. ### Step 1: Understand the problem We have a cliff that is 100 meters high. A man at the top of the cliff observes two points on the ground at angles of depression of 30° and 60°. We need to find the horizontal distances from the base of the cliff to these two points and then calculate the difference between these distances. ### Step 2: Draw a diagram Draw a vertical line representing the cliff and label its height as 100 meters. From the top of the cliff, draw two lines downwards at angles of 30° and 60° to represent the lines of sight to the two points on the ground. Label the horizontal distances from the base of the cliff to the two points as x (for the point at 30°) and y (for the point at 60°). ### Step 3: Use the tangent function For the angle of depression, we can use the tangent function, which relates the opposite side (height of the cliff) to the adjacent side (horizontal distance). 1. For the angle of depression of 30°: \[ \tan(30°) = \frac{\text{Height of cliff}}{x} = \frac{100}{x} \] We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\). Therefore: \[ \frac{1}{\sqrt{3}} = \frac{100}{x} \] Rearranging gives: \[ x = 100 \sqrt{3} \] 2. For the angle of depression of 60°: \[ \tan(60°) = \frac{\text{Height of cliff}}{y} = \frac{100}{y} \] We know that \(\tan(60°) = \sqrt{3}\). Therefore: \[ \sqrt{3} = \frac{100}{y} \] Rearranging gives: \[ y = \frac{100}{\sqrt{3}} \] ### Step 4: Find the difference between the two points Now that we have both distances, we can find the difference between them: \[ \text{Difference} = x - y = 100 \sqrt{3} - \frac{100}{\sqrt{3}} \] ### Step 5: Simplify the expression To simplify the expression, we can find a common denominator: \[ \text{Difference} = 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{100 \cdot 3}{\sqrt{3}} - \frac{100}{\sqrt{3}} = \frac{300 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} \] ### Final Answer Thus, the difference between the two points is: \[ \frac{200}{\sqrt{3}} \text{ meters} \]
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FIITJEE-HEIGHTS & DISTANCE -ASSIGNMENT PROBLEMS (OBJECTIVE) LEVEL-II
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  2. The height of a house subtends a right angle at the opposite street li...

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  4. The angle of elevation of a tower PQ at a point A due north of it is 3...

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  5. An electric pole stands at the vertex A of the equilateral triangular ...

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  6. A pole on the ground lenses 60^(@) to the vertical. At a point a meter...

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  7. A 10 meters high tower is standing at the centre of an equilateral tri...

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  8. A tower subtends an angle alpha at a point A in the plane of its base ...

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  9. A tower subtends angle theta, 2theta, 3theta at the three points A,B a...

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  10. A spherical balloon of radius 50 cm, subtends an angle of 60^(@) at a ...

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  11. PQ is a post of height a, AB is a tower of height h at a distance x fr...

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  12. The angle of elevation of the top of a hill from a point is alpha. Aft...

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  13. The base of a cliff is circular. From the extremities of a diameter of...

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  14. A round balloon of radius r subtends an angle alpha at the eye of the ...

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  15. .A house of height 100 m substends a right angle at the window of an o...

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  16. A tower subtends an angle alpha at a point A in the plane of its...

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  17. ABCD is a rectangular field. A vertical lamp post of height 12 m stand...

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  18. A tower of x metres height has flag staff at its top. The tower and th...

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  19. An aeroplane flying horizontally , 1km above the ground , is observed...

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  20. A man in a boat rowed away from a cliff 150 m high takes 2 min, to cha...

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