Write a function and solve for the

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago

Please answer this math question for me

vampirchik [111]

Answer:

D=90° E=103° F=77°

Step-by-step explanation:

D+90°=180°

D=90°

F+103°=180°

F=77°

E + F=180°

E=103°

6 0

3 years ago

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In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. In the same year

LenaWriter [7]

Answer:

Step-by-step explanation:

Let x represent the number of years it will take the two colleges to have the same enrollment.

In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. This means that the expected number of students at college A in x years time is

12900 + 900x

In the same year, there were 25,000 students at college B, with a projected enrollment decline of 700 students per year. This means that the expected number of students at college B in x years time is

25000 - 700x

For both colleges to have the same enrollment,

12900 + 900x = 25000 - 700x

900x + 700x = 25000 - 12900

1600x = 12100

x = 12100/1600

x = 7.56

Approximately 8 years

The year would be 2000 + 8 = 2008

4 0

3 years ago

Which expression is equivalent to 6⋅6⋅6? 36 6³ 66

devlian [24]

Answer:

6^3

Step-by-step explanation:

6^3 is the same as 6*6*6

5 0

2 years ago

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PLEASE HELP WILL MARK BRAINLIEST Question 8

mote1985 [20]

Answer:

Step-by-step explanation:

a)y=-2x+8

y=4x-7

solution

b)y=x-4

y=3x-2

solution

c)x – 3y = -3

x – 3y = 6

no solution

d)3x +y = 3

6x + 2y = 6

infinite.

5 0

3 years ago

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