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
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brainly.com/question/14350740
A statement that has mathematical expressions that are equal.
Examples:
2+1=3
11x4=44
Answer:
m∠Q ≈ 53°
Step-by-step explanation:
To find the measure of ∠Q, the law of cosines will need to be used. Lowercase letters represent the side lengths, while upper case letters represent angles.
In this situation, 'A' will be ∠Q. Therefore:
17² = 18² + 20² -2(18)(20)cosQ
Simplify:
289 = 324 + 400 -2(360)cosQ
Continue simplifying down:
-435 = -720cosQ
Divide both sides by '-720':
0.604 = cosQ
m∠Q ≈ 52.83 or 53° rounded to the nearest whole degree.
