Home
Class 10
MATHS
A kite is flying at a height of 50sqrt(3...

A kite is flying at a height of `50sqrt(3)` m from the horizontal. It is attached with a string and makes an angle `60^(@)` with the horizontal. Find the length of the string.

Text Solution

AI Generated Solution

The correct Answer is:
To find the length of the string attached to the kite, we can use the properties of a right triangle formed by the height of the kite, the horizontal distance, and the string itself. Here’s the step-by-step solution: ### Step 1: Understand the Problem We know that the kite is flying at a height of \(50\sqrt{3}\) meters from the horizontal and makes an angle of \(60^\circ\) with the horizontal. We need to find the length of the string, which acts as the hypotenuse of the right triangle formed. ### Step 2: Draw the Right Triangle 1. Draw a horizontal line to represent the ground. 2. From a point on this line, draw a vertical line up to a height of \(50\sqrt{3}\) meters. This represents the height of the kite. 3. Connect the top of this vertical line to the point on the horizontal line where the string is attached, forming a right triangle. ### Step 3: Label the Triangle - Let the point where the kite is located be point A (top of the vertical line). - Let the point on the ground directly below the kite be point B. - Let the point where the string is attached to the ground be point C. - The height \(AB = 50\sqrt{3}\) m. - The angle \(CAB = 60^\circ\). - The length of the string \(AC\) is what we need to find. ### Step 4: Use the Sine Function In the right triangle \(ABC\): - The sine of angle \(C\) (which is \(60^\circ\)) is given by the formula: \[ \sin(C) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AB}{AC} \] Here, \(AB\) is the height \(50\sqrt{3}\) m and \(AC\) is the length of the string \(x\). ### Step 5: Set Up the Equation Substituting the known values into the sine formula: \[ \sin(60^\circ) = \frac{50\sqrt{3}}{x} \] We know that \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\). ### Step 6: Solve for \(x\) Substituting \(\sin(60^\circ)\) into the equation: \[ \frac{\sqrt{3}}{2} = \frac{50\sqrt{3}}{x} \] Cross-multiplying gives: \[ \sqrt{3} \cdot x = 100\sqrt{3} \] Dividing both sides by \(\sqrt{3}\): \[ x = 100 \] ### Final Answer The length of the string is \(100\) meters. ---
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • SOME APPLICATIONS OF TRIGONOMETRY

    CBSE COMPLEMENTARY MATERIAL|Exercise SHORT ANSWER TYPE QUESTIONS|20 Videos
  • SOME APPLICATIONS OF TRIGONOMETRY

    CBSE COMPLEMENTARY MATERIAL|Exercise LONG ANSWER TYPE QUESTIONS|19 Videos
  • REAL NUMBERS

    CBSE COMPLEMENTARY MATERIAL|Exercise LONG ANSWER TYPE QUESTION|15 Videos
  • STATISTICS

    CBSE COMPLEMENTARY MATERIAL|Exercise Practice -Test|5 Videos

Similar Questions

Explore conceptually related problems

A kite is flying at a height of 75 metres from the ground level,attached to a string inclined at 60o to the horizontal.Find the length of the string to the nearest metre.

The string of a kite is 150 m long and it makes an angle 60^(@) with the horizontal. Find the height of the kite above the ground. (Assume string to be tight)

Knowledge Check

  • A kite is flying at a height of 123 m. The thread attached to it is assumed to bestretched straight and makes an angle of 60 degree with the level ground. The length ofthe string is (nearest to a whole number):

    A
    140 m
    B
    139 m
    C
    142 m
    D
    138 m
  • A kite is flown with a thread of 250 m length. If the thread is assumed to be stretched and makes an angle of 60^(@) with the horizontal, then the height of the kite above the ground is (approx.) :

    A
    216.25 m
    B
    215.25 m
    C
    212.25 m
    D
    210.25 m
  • The string of a kite is 150 m long and it makes an angle of 60^(@) with the horizontal. The height f the kite from the ground is

    A
    75 m
    B
    `75sqrt3` m
    C
    80 m
    D
    `80sqrt3m`
  • Similar Questions

    Explore conceptually related problems

    A kite if flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground . If the length of the string is 40 sqrt(3) . find the inclination of the string with the ground

    A body of mass mis suspended by two strings making angles alpha and beta with the horizontal. Find the tensions in the strings.

    A block A of mass 5 sqrt( 4) m rests on a rough horizontal plane relative to which its coefficient of friction is mu A light string attached to this block passes over a light frictionless pulley and another block B of mass m hangs from it. It is observed that the block A just slips when the string makes an angle of 60^(@) with the horizontal . Find the coefficient of static friction between the block A and the plane. If the same force acting through the sting is applied to the block A in the horizontal direction, find the acceleration of the block A.

    A kite is flying at a height 80 m above the ground . The string of the kite which is temporarily attached to the ground makes an angle 45^(@) with the ground. If there is no slack in the string, then the length of the string is

    A kite is flying at the height of 75m from the ground. The string makes an angle theta (where cot theta = 8/15 ) with the level ground. Assuming that there is no slack in the string the new length of the string is equal to :