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In order to estimate the height of a building, two students stand at a certain distance from the building at street level. From this point, they find the angle of elevation from the street to the top of the building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height of the building to the nearest foot.

Respuesta :

Answer:

758 ft

Step-by-step explanation:

Note: Refer to the figure

Now from the figure:

In Δ AOD

tan 39° = H/(300 + x)      .................(1)

and in Δ BOD

tan50° = H/x   .................(2)

on dividing 1 by 2 we get

0.6794 = x/(300 + x)

or

203.84 + 0.6794x = x

or

0.3206x = 203.84

or

x = 635.80 ft

now using the equation (2)

we get

tan 50° = H/x

or

H = xtan50°

on substituting the value of x, we get

H = 635.80 × tan 50

or

H = 757.72 ft ≈ 758 ft

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The height of the building to the nearest foot is 758 feet

Find the diagram of the given question attached

Using the SOH CAH TOA identity on triangle OBD and OAD

In triangle OAD

[tex]tan \angle A = \frac{OD}{OA}\\tan 39^0=\frac{H}{300+x}\\[/tex]

For the triangle OBD

[tex]tan \angle B = \frac{OD}{OB}\\tan 50^0=\frac{H}{x}\\[/tex]

Divide both resulting equations to get the value of "x"

[tex]\frac{tan39}{tan50}= \frac{x}{300+x}\\\frac{0.8098}{1.1918} = \frac{x}{300+x}\\0.6795 = \frac{x}{300+x}\\0.6795(300+x) = x\\203.84 + 0.6795x = x\\0.6795x - x = -203.84\\0.3205x = 203.84\\x = \frac{203.84}{0.3205} \\x= 636.00feet[/tex]

Next is to get the height of the building to the nearest foot.

Recall that [tex]tan 50^0=\frac{H}{x}\\[/tex]

H = xtan50

H = 636tan50

H = 636(1.1918)

H = 757.96 feet

Hence the height of the building to the nearest foot is 758 feet

Learn more here: https://brainly.com/question/13038716

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