A 32 foot tall tree casts a shadow that is 48 feet long How far away from the tree is a 6 ft person

Mathematics

1 answer:

dedylja [7]3 years ago

7 0

Answer:

39 feet

Step-by-step explanation:

In the attached diagram

The height of the tree = AB

The height of the man =DE

The length of the shadow =BC

We want to determine how far away from the tree the person is, i.e. |AD|=y

In Triangle ABC

$Tan \theta=\dfrac{32}{48}$

In Triangle CED

$Tan \theta=\dfrac{6}{48-y}$

Therefore:

$Tan \theta=\dfrac{32}{48}=\dfrac{6}{48-y}\\\dfrac{32}{48}=\dfrac{6}{48-y}\\$Cross Multiply\\32(48-y)=48*6\\1536-32y=288\\32y=1536-288\\32y=1248\\Divide both sides by 32\\y=39$

Therefore, the distance of the man from the tree is 39 feet.

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Solve the System by Graphing:

Marat540 [252]

Answer:

B) One solution
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The graph is below.

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Explanation:

I used GeoGebra to graph the two lines. Desmos is another free tool you can use. There are other graphing calculators out there to choose from as well.

Once you have the two lines graphed, notice that they cross at (2, -2) which is where the solution is located. This point is on both lines, so it satisfies both equations simultaneously. There's only one such intersection point, so there's only one solution.

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To graph these equations by hand, plug in various x values to find corresponding y values. For instance, if you plugged in x = 0 into the first equation, then,

y = (-3/2)x+1

y = (-3/2)*0+1

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The point (0,1) is on the first line. The point (2,-2) is also on this line. Draw a straight line through the two points to finish that equation. The other equation is handled in a similar fashion.