Problem 25 - Algebra 2 and Trigonometry

Question

A pole perpendicular to the ground is braced by a wire 13 feet long that is fastened to the ground 5 feet from the base of the pole. The measure of the angle the wire makes with the ground is \(\theta .\) Find the value of: $$\begin{array}{llll}{\text { a. } \sec \theta} & {\text { b. } \csc \theta} & {\text { c. } \cot \theta}\end{array}$$

Step-by-Step Solution

Answer

\( \sec \theta = \frac{13}{5}, \csc \theta = \frac{13}{12}, \cot \theta = \frac{5}{12} \)

Step 1: Understand the problem

First, we identify the right triangle formed by the wire, the ground, and the pole. The wire is the hypotenuse, measuring 13 feet. The ground is a leg, measuring 5 feet from the pole's base to where the wire touches the ground.

Step 2: Use the Pythagorean Theorem

The hypotenuse (the wire) is 13 feet and one leg (the ground distance) is 5 feet. We can find the other leg (the pole's height) using the Pythagorean Theorem: \[ c^2 = a^2 + b^2 \]where \(c = 13\), \(a = 5\), and \(b\) (the pole's height) is unknown. Substituting in, we have:\[ 13^2 = 5^2 + b^2 \]

Step 3: Solve for the pole's height using Pythagorean Theorem

Calculate the pole's height:\[ 169 = 25 + b^2 \]Subtract 25 from both sides:\[ 144 = b^2 \]Taking the square root, we find:\[ b = 12 \]Thus, the pole's height is 12 feet.

Step 4: Calculate \( \sec \theta \)

Recall that \( \sec \theta \) is the reciprocal of \( \cos \theta \), and \( \cos \theta = \frac{adjacent}{hypotenuse} \). Here, \( cos\theta = \frac{5}{13} \), thus:\[ \sec \theta = \frac{1}{\cos \theta} = \frac{13}{5} \]

Step 5: Calculate \( \csc \theta \)

\( \csc \theta \) is the reciprocal of \( \sin \theta \), and \( \sin \theta = \frac{opposite}{hypotenuse} \). Here, \( sin\theta = \frac{12}{13} \), hence:\[ \csc \theta = \frac{1}{\sin \theta} = \frac{13}{12} \]

Step 6: Calculate \( \cot \theta \)

\( \cot \theta \) is the reciprocal of \( \tan \theta \), and \( \tan \theta = \frac{opposite}{adjacent} \). Here, \( tan\theta = \frac{12}{5} \), so:\[ \cot \theta = \frac{1}{\tan \theta} = \frac{5}{12} \]

Key Concepts

Right Triangle

The concept of a right triangle is crucial for understanding problems like the one presented. A right triangle is a type of triangle that has one of its angles measuring exactly 90 degrees. This is why it's called a "right angle". The side opposite the right angle is known as the hypotenuse and is the longest side of the triangle. The other two sides are known as the "legs".
In our problem, the setup forms a right triangle. The hypotenuse is the wire measuring 13 feet. The ground and the pole act as the two legs. With the ground at 5 feet long, it completes the right angle with the upright pole. This type of setup allows us to use key mathematical tools like the Pythagorean Theorem and trigonometric ratios to solve for unknown values.
Remember that identifying the hypotenuse is a critical first step. Knowing which side is the hypotenuse helps in applying other trigonometric properties. This foundation sets us up to easily compute and use ratios that relate different sides of the triangle to the angles, like what we'll explore next.

Pythagorean Theorem

The Pythagorean Theorem is a powerful rule used to find a missing length in right triangles. The theorem states that in a right triangle, the square of the hypotenuse length is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it's written as:
\[ c^2 = a^2 + b^2 \]
where \(c\) represents the hypotenuse while \(a\) and \(b\) represent the triangle's legs.
In our exercise, suppose we're given the hypotenuse (wire) as 13 feet and one leg (ground) as 5 feet. Applying the Pythagorean Theorem finds the missing side, the height of the pole:

• Firstly, substitute the known lengths into the equation: \(13^2 = 5^2 + b^2\)
• Calculate to find the unknown leg: \[169 = 25 + b^2\]
• Solve for \(b^2\): \(b^2 = 144\)
• Take the square root: \(b = 12\)

Thus, the pole is 12 feet tall. This calculation completes the triangle, providing all side lengths needed for further trigonometric calculations.

Trigonometric Ratios

Trigonometric ratios relate the angles of a right triangle to the lengths of its sides. These are important for solving for angles or side lengths in triangles, especially when certain values like the lengths or an angle are given. The main trigonometric ratios include sine \( (\sin)\), cosine \( (\cos)\), and tangent \( (\tan)\).

However, our task explores the reciprocal trigonometric functions: secant \( (\sec)\), cosecant \( (\csc)\), and cotangent \( (\cot)\):

• \(\sec \theta\) is the reciprocal of \(\cos \theta\)
• \(\csc \theta\) is the reciprocal of \(\sin \theta\)
• \(\cot \theta\) is the reciprocal of \(\tan \theta\)

Using this knowledge in our setup:

• For \(\sec \theta\), with known \(\cos \theta = \frac{5}{13}\) (adjacent over hypotenuse), we calculate: \(\sec \theta = \frac{1}{\cos \theta} = \frac{13}{5}\).
• For \(\csc \theta\), as \(\sin \theta = \frac{12}{13}\) (opposite over hypotenuse), we find: \(\csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}\).
• Finally, \(\cot \theta\) with \(\tan \theta = \frac{12}{5}\) (opposite over adjacent) becomes: \(\cot \theta = \frac{1}{\tan \theta} = \frac{5}{12}\).

Each value is derived by relating the triangle's sides to these ratios. This allows us to understand the geometric relationships within the triangle as they relate to its angles.