In traveling across flat land, you see a mountain directly in front of you. Its angle of elevation (to the peak) is 3.5°. After you drive 15 miles closer to the mountain, the angle of elevation is 9° (see figure). Approximate the height of the mountain. (Round your answer to one decimal place.)

In traveling across flat land you see a mountain directly in front of you Its angle of elevation to the peak is 35 After you drive 15 miles closer to the mounta class=

Respuesta :

check the picture below.

[tex]\bf tan(3.5^o)=\cfrac{y}{15+x}\implies 15+x=\cfrac{y}{tan(3.5^o)}\\\\\\ \boxed{x=\cfrac{y}{tan(3.5^o)}-15} \\\\\\ tan(9^o)=\cfrac{y}{x}\implies x\cdot tan(9^o)=y\\\\\\ \left[ \cfrac{y}{tan(3.5^o)}-15 \right]tan(9^o)=y \\\\\\ \left[ \cfrac{y-15tan(3.5^o)}{tan(3.5^o)} \right]tan(9^o)=y \\\\\\[/tex]

[tex]\bf \cfrac{y\cdot tan(9^o)-15tan(3.5^o)tan(9^o)}{tan(3.5^o)}=y \\\\\\ y\cdot tan(9^o)-15tan(3.5^o)tan(9^o)=y\cdot tan(3.5^o) \\\\\\ ytan(9^o)-ytan(3.5^o)=15tan(3.5^o)tan(9^o) \\\\\\ \stackrel{common~factor}{y}[tan(9^o)-tan(3.5^o)]=15tan(3.5^o)tan(9^o) \\\\\\ y=\cfrac{15tan(3.5^o)tan(9^o)}{tan(9^o)-tan(3.5^o)}[/tex]

make sure your calculator is in Degree mode.

the amount is small, but bear in mind that is mile units.
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To solve such problems we must know about Tangent (Tanθ).

What is Tangent (Tanθ) ?

The tangent or tanθ in a right angle triangle is the ratio of its perpendicular to its base. it is given as,

[tex]\rm{Tangent(\theta) = \dfrac{Perpendicular}{Base}[/tex]

where,

θ is the angle,

Perpendicular is the side of the triangle opposite to the angle θ,

The base is the adjacent smaller side of the angle θ.

The height of the mountain is 1.495 miles.

Given to us

  • AB = 15 miles
  • ∠CAD = 3.5°
  • ∠CBD = 9°

In ΔCAD,

[tex]\rm{ tan(\angle A) = \dfrac{Perpendicular}{Base}[/tex]

[tex]tan(\angle A) = \dfrac{CD}{AD}[/tex]

[tex]tan(3.5^o) = \dfrac{CD}{15+x}[/tex]

[tex]CD=tan(3.5^o) \times(x+15)[/tex]

In ΔCBD,

[tex]\rm{ tan(\angle B) = \dfrac{Perpendicular}{Base}[/tex]

[tex]tan(\angle B) = \dfrac{CD}{BD}[/tex]

[tex]tan(9^o) = \dfrac{CD}{x}[/tex]

[tex]CD=tan(9^o) \times(x)[/tex]

Equating CD,

[tex]tan(3.5^o) \times (x+15) = tan(9^o) \times x\\\\(x+15)= \dfrac{tan(9^o) }{tan(3.5^o) }\times x\\\\x+15=2.59\ x\\15 = 2.59x-x\\15= 1.59x\\x = 9.44[/tex]

Substituting the value of x

[tex]CD=tan(9^o) \times(x)[/tex]

[tex]CD=tan(9^o) \times(9.44)\\CD= 1.495[/tex]

Hence, the height of the mountain is 1.495 miles.

Learn more about Tangent (Tanθ) function:

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