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In the figure, a person standing at point A notices that the angle of elevation to the top of the antenna is 47° 30'. A second person standing 40.0 feet farther from the antenna than the person at A finds the angle of elevation to the top of the antenna to be 42° 10'. How far is the person at A from the base of the antenna? (Round your answer to the nearest whole number.)

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The position of the antenna should be let represented as \( B \). The first person is at point \( A \) and the second one is 40 feet farther from the antenna than point \( A \) so let the distance \( AB=x \) feet. Hence the separation between the two individuals or the distance from the antenna to the second person (at \( C \)) is \( x + 40 \) feet.

Starting from point \( A \), the angle of elevation to the top of the antenna \( \angle ABD = 47^\circ 30’ \).

From point \( C \), the angle of elevation to the top of the antenna \( \angle CBD = 42^\circ 10’ \).

First, convert the angles to decimal degrees:First, convert the angles to decimal degrees:

[ Angle of the triangle ABD = 47. ] 5^\circ \]

They found that \( \angle CBD = 42\). 1667^\circ \]

Let \( h \) represent the height of the antenna.

Using trigonometry in triangle \( ABD \):Using trigonometry in triangle \( ABD \):

The tangent of angle ABD is equal to the height divided by the segment AB \[ \tan \angle ABD = \frac{h}{AB} \]

\[ \tan 47. 5^\circ = \frac{h}{x} \]

Here the equation is [ h = x \cdot \tan 47. 5^\circ ], where h is the height and x could be either the length of the ramp or the distance between the bottom the ramp and the object.

Using trigonometry in triangle \( CBD \):Using trigonometry in triangle \( CBD \):

\[ \tan \angle CBD= \frac{h}{BC} \]

\[ \tan 42. We have \[ 1667^\circ = \frac{h}{x + 40} \]

h = (x + 40)* tan 42. 1667^\circ \]

Since both expressions represent \( h \), equate them:Since both expressions represent \( h \), equate them:

Equating the values of tangent we have: \[ x \cdot \tan 47. 5^\circ = (x + 40) \cdot \tan 42. 1667^\circ \]

Now, solve for \( x \):Now, solve for \( x \):

That is,_UT sine  = UT cos (90 - 47. 5) _x tan 47. 5 = x tan 42. So substituting 1667º by tan(1667º) gives: \[1667^\circ + 40 \cdot \tan 42. 1667^\circ \]

[ x \cdot (\tan 47 5^\circ - \tan 42. Thus by using the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1- \tan A \tan B}\) we have \[(180 - 1667)^\circ = 40 \cdot \tan 42. \] 1667^\circ \]

[ To solve nth term, x equates to forty times tangent of forty two point one hundred and sixty seven times the tangent of forty seven point ] 5^\circ - \tan 42. 1667^\circ} \]

Calculate \( x \):

\[ \tan 47. 5^\circ \approx 1. 144 \]

\[ \tan 42. 1667^\circ \approx 0. 911 \]

Answer:

Step-by-step explanation:

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