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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In three dimensions, the cross product of two vectors is defined as shown below

Then, solving the determinant

In our case,

Where we used the formula for AxB to calculate ixj.
Finally,

Thus, (i+j)x(ixj)=i-j
The probability of occurrence for the events A, B and C is; 1/4.
<h3>What is the probability of occurrence of.the described events?</h3>
For the first event A in which case, there's no odd number on the first two rolls, the possible events are; EEE and EEO. Consequently, the required probability is;
Event A = 2/8 = 1/4.
For the event B in which case, there's an even number on both the first and last rolls; the possible events are; EEE and EOE. Consequently, the required probability is;
Event B = 2/8 = 1/4.
For the event C in which case, there's an odd number on each of the first two rolls; the possible events are; OOO and OOE. Consequently, the required probability is;
Event C = 2/8 = 1/4.
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