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A bee is on a hill looking at a building. The building is 400 feet tall. The angle of elevation from the bee to the top of the building is \(\color[rgb]{0.35,0.35,0.35}4^\circ\) and the angle of depression from the bee to the bottom of the building is \(\color[rgb]{0.35,0.35,0.35}2^\circ.\) What is the shortest distance the bee will have to fly to reach the building?

 May 14, 2021
edited by Guest  May 14, 2021
 #1
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Let $X$ be the position of the be.

Let $A$ be the highest point of the building.

Let $B$ be the lowest point of the building.

Let $C$ be the point that has the same altitude to $X$ (aka the shortest distance, which is what we are trying to find.)

 

We are given $AB = 400.$

Since we are given two angles and $AB$, we will try to relate $AB$ to $XC.$

Note $AB = AC + BC.$

We have that $AC = XC \tan 4^{\circ} \cong 0.070 XC.$

We also have that $BC = XC \tan 2^{\circ} \cong 0.035 XC.$

Thus we have $AB \cong 0.105 XC.$

And since we also have $AB = 400,$ we have that $0.105 XC \cong 400.$

$XC \cong \frac{200}{21} \cdot 400 \cong \boxed{3800 \text{ feet.}}$

 May 14, 2021
 #2
avatar+32762 
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See image below

 

d ( tan 2 + tan4 ) = 400

d = 3815 ft

 

 

 May 14, 2021

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