A fireman spots a stranded woman in the window of a 200-foot tall office building. From the
point where the fireman is standing, the angle of elevation to the stranded woman is 42° 33'
and to the top of the building is 58° 6'. How far from the ground is the stranded woman?

Answer :

lublana

Answer:

114.3 foot

Step-by-step explanation:

Height of building,h=200 foot

Let BC=x

[tex]tan\theta=\frac{perpendicular\;side}{base}[/tex]

Using the formula

[tex]tan(58^{\circ} 6')=\frac{200}{CD}[/tex]

[tex]CD=\frac{200}{tan(58^{\circ} 6')}[/tex]

[tex]CD=124.5 foot[/tex]

Now,

[tex]\frac{BC}{CD}=tan(42^{\circ}33')[/tex]

[tex]\frac{x}{124.5}=0.9179[/tex]

[tex]x=124.5\times 0.9179[/tex]

[tex]x=114.3 foot[/tex]

Hence, the distance of woman from the  ground=114.3 foot

The ground is 123.24 ft far from the stranded woman.

Given the total height of the office building is 200 ft.

And the angle of elevation of the top of the building is 58° 6' and the angle of elevation to the stranded woman is 42° 33'.

We know that, from trigonometric ratios,[tex]tan \theta =\frac{perpendicular}{base}[/tex]

here [tex]\theta[/tex] is the angle of elevation of the top of the building from the fireman.

Now, [tex]tan ( 58^{0} 6')[/tex][tex]=\frac{200}{base}[/tex]

[tex]1.488=\frac{200}{base}[/tex]

[tex]base=\frac{200}{1.488}[/tex]

[tex]base=134.40[/tex] ft

Now the woman is stranded in the window of the building say at the height of h ft, and the angle of elevation of the woman from the fireman is  42° 33'.

So, [tex]tan (42^{0} 33')[/tex][tex]= \frac{perpendicular}{base}[/tex]

[tex]tan(42^{0} 33' )= \frac{h}{134.40}[/tex]

[tex]0.917=\frac{h}{134.40}[/tex]

[tex]0.917\times134.40=h[/tex]

[tex]h=123.24 ft[/tex]

Hence the ground is 123.24 ft far from the stranded woman.

For more details on angle of elevation follow the link:

https://brainly.com/question/12483071

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