use a sketch of a reference triangle to rewrite the expression as an algebraic expression in x: sin(tan-1x)?
Transcribed Image Text:### Trigonometric Expressions and Applications
**Problem Statement:**
1. Use a sketch of a reference triangle to rewrite the expression as an algebraic expression in \( x \):
\[
\sin(\tan^{-1}(x))
\]
2. A 20-ft pole casts a shadow as shown in the figure. Find the angle of elevation of the sun when the shadow is \( s = 35 \) ft long. (Round your answer to one decimal place.)
**Diagram Explanation:**
The diagram in the image is a right-angled triangle labeled as follows:
- Angle \( A \) at the point \( A \).
- Side \( B \) is opposite to angle \( C \).
- Side \( C \) is adjacent to angle \( B \).
For the first part of the problem:
To rewrite \( \sin(\tan^{-1}(x)) \) using a reference triangle:
1. Draw a right triangle where the angle \( \theta \) satisfies \(\tan(\theta) = x\). This implies that opposite side \( = x \) and adjacent side \( = 1 \).
2. By the Pythagorean theorem, the hypotenuse \( = \sqrt{1^2 + x^2} = \sqrt{1 + x^2} \).
3. Therefore, \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1 + x^2}}\).
Thus, \(\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}}\).
For the second part of the problem:
To find the angle of elevation of the sun:
1. Draw a right triangle where one leg (the height of the pole) is 20 ft and the other leg (the length of the shadow) is 35 ft.
2. Use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{20}{35} = \frac{4}{7}
\]
3. Solve for \(\theta\):
\[
\theta = \tan^{-1}\left(\frac{4}{7}\right)
\]
4. Using a calculator,
\[
\theta \approx 29.7^\circ
\
Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).
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![### Trigonometric Expressions and Applications
**Problem Statement:**
1. Use a sketch of a reference triangle to rewrite the expression as an algebraic expression in \( x \):
\[
\sin(\tan^{-1}(x))
\]
2. A 20-ft pole casts a shadow as shown in the figure. Find the angle of elevation of the sun when the shadow is \( s = 35 \) ft long. (Round your answer to one decimal place.)
**Diagram Explanation:**
The diagram in the image is a right-angled triangle labeled as follows:
- Angle \( A \) at the point \( A \).
- Side \( B \) is opposite to angle \( C \).
- Side \( C \) is adjacent to angle \( B \).
For the first part of the problem:
To rewrite \( \sin(\tan^{-1}(x)) \) using a reference triangle:
1. Draw a right triangle where the angle \( \theta \) satisfies \(\tan(\theta) = x\). This implies that opposite side \( = x \) and adjacent side \( = 1 \).
2. By the Pythagorean theorem, the hypotenuse \( = \sqrt{1^2 + x^2} = \sqrt{1 + x^2} \).
3. Therefore, \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{1 + x^2}}\).
Thus, \(\sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}}\).
For the second part of the problem:
To find the angle of elevation of the sun:
1. Draw a right triangle where one leg (the height of the pole) is 20 ft and the other leg (the length of the shadow) is 35 ft.
2. Use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{20}{35} = \frac{4}{7}
\]
3. Solve for \(\theta\):
\[
\theta = \tan^{-1}\left(\frac{4}{7}\right)
\]
4. Using a calculator,
\[
\theta \approx 29.7^\circ
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