Question

I need help answering this practice problem from my trip prep book

Answer 1

From the statement of the problem, we know that:

• the height of the tree is 80ft,

,

• the first angle of inclination is 68°,

,

• the second angle of inclination is 41°.

We define:

• d as the distance for the first angle of inclination,

,

• x as the distance that Corey walk to the position for the second angle of inclination.

Using the data of the problem, we make the following graph:

From the triangle, we see two triangles:

1) △BTA1, with:

• θ = angle = ,68°,.

,

• OS = opposite side to the angle = 80ft,

,

• AS = adjacent side to the angle = ,d,,

2) △BTA2, with:

• θ = angle = ,41°,.

,

• OS = opposite side to the angle = 80ft,

,

• AS = adjacent side to the angle =, x + d,,

Now from trigonometry, we have the following relation:

`\tan \theta=(OS)/(AS)\text{.}`

Using this equation for each triangle, we get the following equations:

`\begin{gathered} \tan 68^(\circ)=(80ft)/(d), \\ \tan 41^(\circ)=(80ft)/(x+d)\text{.} \end{gathered}`

From the first equation, we get the value of d:

`\begin{gathered} \tan 68^(\circ)=(80ft)/(d), \\ d=(80ft)/(\tan68^(\circ)). \end{gathered}`

Solving the second equation for x and replacing the value of d, we get:

`\begin{gathered} \tan 41^(\circ)=(80ft)/(x+d), \\ x=(80ft)/(\tan41^(\circ))-d, \\ x=(80ft)/(\tan41^(\circ))-(80ft)/(\tan68^(\circ)), \\ x\cong59.71ft. \end{gathered}`

Answer

Corey stepped back 59.71ft to gain a better view of the bird.