Question

NO LINKS!!! Part 11
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Answer 1

Answer:

  39.4 feet

Step-by-step explanation:

The tangent trig relation can be used to relate the height of the tree to the distance from the observation point. We can start with the tangent relation, then apply it to both observation points.

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Let h represent the height of the tree. Then for some distance d to the tree, we have ...

  Tan = Opposite/Adjacent

  tan(30°) = h/d

  tan(20°) = h/(d+40)

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Solving the first equation for d and substituting into the second equation gives ...

  d = h/tan(30°)

  tan(20°) = h/(h/tan(30°) +40) . . . . . substitute for d

  h/tan(30°) +40 = h/tan(20°) . . . . . . multiply by (h/tan(30°)+40)/tan(20°)

  h(1/tan(20°) -1/tan(30°) = 40 . . . . . subtract h/tan(30°) and factor

  h = 40tan(20°)tan(30°)/(tan(30°) -tan(20°)

  h ≈ 39.392 . . . feet

The height of the tree is about 39.4 feet.

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As you can see from the attachment, a graphing calculator can solve these equations easily.

Answer 2

Answer:

39.4 ft (nearest tenth)

Step-by-step explanation:

Create two equations using trig ratios and the information given (refer to the attached diagram).  Equate the equations and solve.

First equation

Let x = distance along the path with 30° angle of elevation

Let h = height of the tree

Using the tangent trig ratio:

$\mathsf{\tan(\theta)=\dfrac{opposite \ side}{adjacent \ side}}$

Given:

• angle = 30°
• side opposite the angle = h
• side adjacent the angle = x

$\implies \mathsf{\tan(30)=\dfrac{h}{x}}$

Rearrange to make x the subject:

$\implies \mathsf{x=\dfrac{h}{\tan(30)}}$

$\implies \mathsf{x=\sqrt{3} \ h}$

Second equation

Distance along the path with 20° angle of elevation = x + 40

Let h = height of the tree

Using the tangent trig ratio:

$\mathsf{\tan(\theta)=\dfrac{opposite \ side}{adjacent \ side}}$

Given:

• angle = 20°
• side opposite the angle = h
• side adjacent the angle = x + 40

$\implies \mathsf{\tan(20)=\dfrac{h}{x+40}}$

Rearrange to make x the subject:

$\implies \mathsf{x=\dfrac{h}{\tan(20)}-40}$

Now equate the 2 equations and solve for h:

$\implies \mathsf{\sqrt{3} \ h=\dfrac{h}{\tan(20)}-40}}$

$\implies \mathsf{\dfrac{h}{\tan(20)}-\sqrt{3} \ h=40}}$

$\implies \mathsf{\dfrac{h-\sqrt{3} \ h\tan(20)}{\tan(20)}=40}}$

$\implies \mathsf{h-\sqrt{3} \ h\tan(20)=40{\tan(20)}}$

$\implies \mathsf{h(1-\sqrt{3} \tan(20))=40{\tan(20)}}$

$\implies \mathsf{h=\dfrac{40{\tan(20)}}{1-\sqrt{3} \tan(20)}}$

$\implies \mathsf{h=39.39231012...}$

Therefore, the height of the tree is 39.4 ft (nearest tenth)