The height of the kite above the ground is 58.68 ft
Let x be the height of the kite above Chee's hand.
The height of the kite above Chee's hand, the string and the horizontal distance between Chee and the kite form a right angled triangle with hypotenuse side, the length of the string and opposite side the height of the kite above Chee's hand.
Since we have the angle of elevation from her hand to the kite is 29°, and the length of the string is 100 ft.
From trigonometric ratios, we have
tan29° = x/100
So, x = 100tan29°
x = 100 × 0.5543
x = 55.43 ft.
Since Chee's hand is y = 3.25 ft above the ground, the height of the kite above the ground, L = x + y
= 55.43 ft + 3.25 ft
= 58.68 ft to the nearest hundredth of a foot
So, the height of the kite above the ground is 58.68 ft
Learn more about height of kite above the ground here:
brainly.com/question/14350740
Answer:
Step-by-step explanation:
In a static position, the sum of the forces is always 0.
so:
F1 = 480 N
F2 = -600 N
F3 = ?
F1 + F2 + F3 = 0
480 - 600 + F3 = 0
F3 = 120 N (the answer is positive, so the force acts in the same direction as F1)
Answer:
f(x) = 2x + 2^4 is a function.
Triangle ABD is i<span>nscribed in circle and </span>has hypotenuse as circle's diameter so angle ABD must be 90°
Since it is parallelogram, ∠ABD = ∠BDC = 90° and it is given that ∠ADB = 30°
So we have ∠CDO = ∠CDB + ∠ADB = 30° + 90° = 120°
Final answers: ∠CDO = 120° and ∠ABD = 90°
Its B bro becauseiys a negative so you would get rid of the symbol then flip the negative
