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dybincka [34]
3 years ago
13

Corey spies a bald eagle in a tall tree. He estimates the height of the tree to be 60 feet and the angle of elevation to the bir

d from where he stands to be 64°. The leaves on the tree make it difficult for Corey to watch the bird, so he takes many steps away from the tree to get a better view. He now estimates his angle of elevation to be 37°.
How many feet did Corey step back to gain a better view of the bird? Round your answer to the nearest hundredth of a foot.
Mathematics
1 answer:
pogonyaev3 years ago
4 0

Answer:

Corey stepped back 59.71 feet

Step-by-step explanation:

Right Triangles

They are a special type of triangle where one of its internal angles is 90°. The basic trigonometric equations stand in this type of triangles. If x is the adjacent leg of a given angle , and y is the opposite leg to the same angle, then :

tan0= \frac{y}{x}

We can solve for x to get

x = \frac{y}{tan0}

Corey estimates the height of the tree to be 80 feet. This means we know y=80 and . Let's find the horizontal distance at which Corey is looking at the bald eagle:

x1 =\frac{80}{tan68}

x1 = 32.32 feet

Now Corey moves back to watch the very same tree at an elevation angle of 41°. The tree has the same height, so

x2 =\frac{80}{tan41}

x2= 93.03 feet

Now we can know the distance Corey walked back by subtracting both distances

<em>correy stepped back 59.71 feet</em>

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\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\  f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2

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We are given the two functions:

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First, find <em>h: </em>

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<em />\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\  & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}

Because (1, -40) and (-1, 24) are points on the graph of <em>h</em>, we have that h(-1) = 40 and h(-1) = 24. In other words:


\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\  -40 &= mn-6m+8n-48  \end{aligned}

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\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24  & = -mn -2m + 8n +16 \end{aligned}

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<em />\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}

Substitute this into one of the two equations above and solve:


\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\  &= (n-2)(n-1) \\ \\ &  \end{aligned}

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\displaystyle n = 2 \text{ or } n = 1

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<em />\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or }  & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}

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