Charlie is watching hot air balloons. Balloon A has risen at a 56° angle. Balloon B has risen at an 81° angle. If the distance f
rom balloon A to the ground is 1,200 feet, how far is balloon B from balloon A? Round your answer to the nearest whole number.
Two points labeled Balloon A and Balloon B are connected to a point labeled Charlie, which is on a straight line labeled Ground. A dashed line connects point Balloon A to line Ground; another dashed line connects point Balloon B to line Ground; both dashed lines form a right angle with line Ground; the angle formed from point Balloon A, point Charlie, and line Ground measures x degrees; and the angle formed by point Balloon B, point Charlie, and line Ground measures y degrees.
999 feet
1,005 feet
1,052 feet
1,102 feet
Two points labeled Balloon A and Balloon B are connected to a point labeled Charlie, which is on a straight line labeled Ground. A dashed line connects point Balloon A to line Ground; another dashed line connects point Balloon B to line Ground; both dashed lines form a right angle with line Ground; the angle formed from point Balloon A, point Charlie, and line Ground measures x degrees; and the angle formed by point Balloon B, point Charlie, and line Ground measures y degrees.
999 feet
1,005 feet
1,052 feet
1,102 feet
2 answers:
Answer:
999 ft
Step-by-step explanation:
In the picture attached, the drawn of the situation is shown
<u>Data</u>
- ∠x = 56°
- ∠y = 81°
- distance from balloon A to the ground is 1,200 feet
From definition:
tan(∠x) = opposite/adjacent
tan(56°) = 1200/adjacent
adjacent to ∠x = 1200/tan(56°)
adjacent to ∠x = 809.41 ft
tan(∠y) = opposite/adjacent
tan(81°) = 1200/adjacent
adjacent to ∠y = 1200/tan(81°)
adjacent to ∠y = 190.06 ft
The distance from balloon A to balloon B is: 809.41 + 190.06 = 999.47 ≈ 999 ft
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Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is . Thus, there are only 35 ways to distribute the blackboards in this case.
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{ ,
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}
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