Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
3. Right Triangles and Trigonometric Ratios
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Exercise 18 Page 924

Practice makes perfect
a We know that the tallest flagpole in the world was in San Francisco. When the angle of elevation of the sun was 55^(∘), the length of the shadow cast by this flagpole was 210 feet.

Notice that we have a right triangle. Then, we can use the trigonometric ratio for tangent to find the height of the flagpole. tan θ = Opposite/Adjacent In our case, the measure of θ is 55^(∘), the length of the adjacent leg is 210, and the length of the opposite leg is x. tan θ = Opposite/Adjacent ⇒ tan 55^(∘) = x/210 Let's solve this equation to find the length of opposite leg!

tan 55 ^(∘) = x/210
210 tan 55^(∘) = x
299.911081 ... = x
300 ≈ x
x ≈ 300

Therefore, the height of the flagpole is about 300 feet.

b We want to find the length of the shadow when the angle of elevation of the sun is 34^(∘). To do so, we will use again the trigonometric ratio for tangent.
tan θ = x/210 This time we will substitute θ= 34^(∘) into this equation and solve it for x. Let's do it!

tan θ = x/210
tan 34^(∘) = x/210
210 tan 34^(∘) = x
141.646788 ... = x
142 ≈ x
x ≈ 142

Therefore, the height of the flagpole will be approximately 142 feet.

c To solve the problems we need to use the trigonometric ratios. But to use them, we need to assume that we have a right triangle formed by the shadow and the flagpole. This is only possible if the flagpole and the shadows are perpendicular to each other.
Flagpole ⊥ Shadow Only with this assumption, we will be able to use trigonometric ratios.