Question 250653: joe is 1.59 m tall, wishes to find the height of a tree. He walks 23.56 m from the base of the tree along the shadow of the tree until his head is in a position where the tip of his shadow exactly overlaps the end of the tree top's shadow. He is now 8.42 m from the end of the shadows. How tall is the tree?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! if I understand this correctly, the scenario is as follows:
A x <--- top of tree
x x
x x
x x
x x
x x D
x x
x x <--- top of joe's head
x x x
x x x
x 1.59m --> x x
x x x
x x x
x x x
B x x x x x x x x x x x x C
|--------23.56m-------|--------8.42m--------|
E
If this is the case, then we can first find the angle DCE and then find the height of the tree.
Tan (DCE) = opposite / adjacent = 1.59 / 8.42 = .188836105
Angle DCE = Arctan (.188836105) = 10.69359057 degrees.
Tan (ACB) = Tan (10.69359057) = .188836105
Tan (ACB) = opposite / adjacent = AB / BC = AB / (23.56 + 8.42) = AB / 31.98
Tan (ACB) = AB / 31.98
Since Tan (ACB) is the same as Tan (DCE), this formula becomes:
.188836105 = AB / 31.98
multiply both sides of this equation by 31.98 to get:
31.98 * .188836105 = AB
AB = 6.038978622 meters high.
Since triangle DCE is similar to triangle ACB, then the corresponding sides are proportional to each other so you could have used a ratio to find the same answer.
The ratio would have been:
DE / CE = AB / CB
This would have come out as:
1.59 / 8.42 = x / 31.98
Cross multiply to get 8.42*x = 1.59 * 31.98
divide both sides by 8.42 to get x = (1.59 * 31.98) / 8.42 = 6.038978622.
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