A child flies a kite at a height of 50 ft, the wind carrying the kite horizontally away from the child at a rate of 26 StartFraction ft Over sec EndFraction . How fast must the child let out the string when the kite is 130 ft away from the child?

24 ft/sec

Step-by-step explanation:

Let the string be the hypotenuse of a right triangle with a constant feet of 50

The vertical leg is considered as the height of the kite

The horizontal leg of the kite is not given, but increases at speed of 26 ft/sec

The length of the leg should be called  leg x , the length of the string L at any point is known by

L² = 50² + x².

The String release rate is given as dL/dt. That is the amount looking for presently   when L = 150. When L = 130, x that is found by substitution

130² = 50² + x², then x = √(130² - 50²) = 120 ft

Applying differentiation method for change of L

2L dL/dt = 0 + 2x dx/dt

When L = 130 and then x = 120

2(130) dL/dt = 2(120) (26)  

dL/dt = 24 ft/sec

according to the triangle sum theorem, all the angles in a triangle must add up to 180 degrees. if we know one angle is already 90, we do 180-90=90. the other 2 angles must add up to 90, exactly.

step-by-step explanation:

hiking down a winding trail.

step-by-step explanation: