Mathematics, 08.07.2019 05:20 parkbro3956

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.)

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Answer from: charlesiarenee0

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,

$\rm{Tangent(\theta) = \dfrac{Perpendicular}{Base}$

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,

$\rm{ tan(\angle A) = \dfrac{Perpendicular}{Base}$

$tan(\angle A) = \dfrac{CD}{AD}$

$tan(3.5^o) = \dfrac{CD}{15+x}$

$CD=tan(3.5^o) \times(x+15)$

In ΔCBD,

$\rm{ tan(\angle B) = \dfrac{Perpendicular}{Base}$

$tan(\angle B) = \dfrac{CD}{BD}$

$tan(9^o) = \dfrac{CD}{x}$

$CD=tan(9^o) \times(x)$

Equating CD,

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

Substituting the value of x

$CD=tan(9^o) \times(x)$

$CD=tan(9^o) \times(9.44)\\CD= 1.495$

Hence, the height of the mountain is 1.495 miles.

Learn more about Tangent (Tanθ) function:

link

In traveling across flat land, you see a mountain directly in front of you. its angle of elevation (

Answer from: Lcorbett7414

Check the picture below.

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

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

make sure your calculator is in Degree mode.

the amount is small, but bear in mind that is mile units.
In traveling across flat land, you see a mountain directly in front of you. its angle of elevation (

Answer from: Quest

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

Answer from: Quest

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p

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