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:
[tex]\tan \theta=\frac{OS}{AS}\text{.}[/tex]
Using this equation for each triangle, we get the following equations:
[tex]\begin{gathered} \tan 68^{\circ}=\frac{80ft}{d}, \\ \tan 41^{\circ}=\frac{80ft}{x+d}\text{.} \end{gathered}[/tex]
From the first equation, we get the value of d:
[tex]\begin{gathered} \tan 68^{\circ}=\frac{80ft}{d}, \\ d=\frac{80ft}{\tan68^{\circ}}. \end{gathered}[/tex]
Solving the second equation for x and replacing the value of d, we get:
[tex]\begin{gathered} \tan 41^{\circ}=\frac{80ft}{x+d}, \\ x=\frac{80ft}{\tan41^{\circ}}-d, \\ x=\frac{80ft}{\tan41^{\circ}}-\frac{80ft}{\tan68^{\circ}}, \\ x\cong59.71ft. \end{gathered}[/tex]
Answer
Corey stepped back 59.71ft to gain a better view of the bird.
