Need help!!! Please
2 answers:
You might be interested in
Observe attached picture.
On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft
When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.
We can apply trigonometry rules to find height of flagpole.
From small triangle containing sign we can find tangens function:
Similarly we can do for large triangle containing flagpole:
We see that these two equations have same left sides. This means that their right sides must also be same:
We can solve for A:
Height of flagpole is 48 feet.
On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft
When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.
We can apply trigonometry rules to find height of flagpole.
From small triangle containing sign we can find tangens function:
Similarly we can do for large triangle containing flagpole:
We see that these two equations have same left sides. This means that their right sides must also be same:
We can solve for A:
Height of flagpole is 48 feet.
2 ways: make a little divide tree based on how many times you can divide it. Like you have sqrt88 you know it can divide by 2. So its 2 and 44, and 44 can be broken to 2 and 22, then 2 and 11. 11 can’t be divided anymore.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.