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The upper part of a tree broken over by the wind makes an angle of `30^(@)` with the ground and the distance of the root from the point where the top touches the ground is 25 m. What was the total height of the tree?

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To solve the problem step by step, we will use trigonometric ratios to find the height of the tree before it was broken. ### Step 1: Understand the problem We have a tree that has been broken by the wind. The upper part of the tree makes an angle of \(30^\circ\) with the ground, and the distance from the root of the tree to the point where the top touches the ground is 25 meters. ### Step 2: Set up the triangle We can visualize this situation as a right triangle: - Let \(x\) be the height of the remaining part of the tree (the part that is still standing). - Let \(y\) be the length of the broken part of the tree (the part that is tilted down). - The distance from the root to the point where the top touches the ground is the base of the triangle, which is 25 meters. ### Step 3: Use the tangent function Using the angle \(30^\circ\), we can apply the tangent function: \[ \tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{25} \] From trigonometric tables, we know that: \[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \] Thus, we can set up the equation: \[ \frac{1}{\sqrt{3}} = \frac{x}{25} \] ### Step 4: Solve for \(x\) To find \(x\), we can rearrange the equation: \[ x = 25 \cdot \frac{1}{\sqrt{3}} = \frac{25}{\sqrt{3}} \text{ meters} \] ### Step 5: Use the cosine function to find \(y\) Next, we will find \(y\) using the cosine function: \[ \cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{25}{y} \] From trigonometric tables, we know that: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] Thus, we can set up the equation: \[ \frac{\sqrt{3}}{2} = \frac{25}{y} \] ### Step 6: Solve for \(y\) Rearranging the equation gives: \[ y = \frac{25 \cdot 2}{\sqrt{3}} = \frac{50}{\sqrt{3}} \text{ meters} \] ### Step 7: Find the total height of the tree The total height of the tree before it was broken is the sum of \(x\) and \(y\): \[ \text{Total height} = x + y = \frac{25}{\sqrt{3}} + \frac{50}{\sqrt{3}} = \frac{75}{\sqrt{3}} \text{ meters} \] ### Step 8: Rationalize the answer To express the answer in a more standard form, we rationalize: \[ \frac{75}{\sqrt{3}} = \frac{75 \cdot \sqrt{3}}{3} = 25\sqrt{3} \text{ meters} \] ### Final Answer The total height of the tree is \(25\sqrt{3}\) meters. ---
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CBSE COMPLEMENTARY MATERIAL-SOME APPLICATIONS OF TRIGONOMETRY-SHORT ANSWER TYPE QUESTIONS
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  2. In the figure, two persons are standing at the opposite direction P & ...

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  3. In the figure, find the value of AB.

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  4. In the figure, find the value of CF.

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  5. If the horizontal distance of the boat from the bridge is 25 m and the...

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  6. State True/False. If the length of the shadow of a tower is increasi...

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  7. If a man standing on a plat form 3 m above the surface of a lake obser...

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  8. The angle of elevation of the top of a tower is 30^(@). If the height ...

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  9. From the top of a hill, the angles of depression of two consecutive ...

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  10. The string of a kite is 150 m long and it makes an angle 60^(@) with t...

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  11. The shadow of a vertical tower on level ground increases by 10 m when ...

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  12. An aeroplane at an altitude of 200 m observes angles of depression of ...

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  13. The angle of elevation of a tower at a point is 45^(@). After going 40...

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  14. The upper part of a tree broken over by the wind makes an angle of 30^...

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  15. A vertical flagstaff stands on a horizontal plane. From a point 100 m ...

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  16. The length of a string between kite and a point on the ground is 90 m....

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  17. An aeroplane, when 3000 m high, passes vertically above anthoer aeropl...

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  18. A 7 m long flagstaff is fixed on the top of a tower on the horizontal...

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  19. From the top of a 7 m high building, the angle of elevation of the top...

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  20. Anand is watching a circus artist climbing a 20m long rope which is ti...

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