Problem 21
Question
A telephone pole is braced by two wires that are both fastened to the ground at a point 15 feet from the base of the pole. The shorter wire is fastened to the pole 15 feet above the ground and the longer wire 20 feet above the ground. a. What is the measure, in degrees, of the angle that the shorter wire makes with the ground? b. Let \(\theta\) be the measure of the angle that the longer wire makes with the ground. Find \(\quad \sin \theta\) and \(\cos \theta .\) c. Find the cosine of the angle between the wires where they meet at the ground. d. Find, to the nearest degree, the measure of the angle between the wires.
Step-by-Step Solution
Verified Answer
a) 45°; b) \(\sin \theta = \frac{4}{5}, \cos \theta = \frac{3}{5}\); c) \(\frac{7\sqrt{2}}{10}\); d) 61°.
1Step 1: Identify Elements and Build a Right Triangle for the Shorter Wire
For the shorter wire, we have a right triangle where the base is 15 feet (distance from the pole to the ground), and the height is 15 feet (the height of the wire on the pole). The wire itself acts as the hypotenuse of this right triangle.
2Step 2: Calculate Angle for Shorter Wire
Using trigonometry, we find the angle \(\alpha\) that the shorter wire makes with the ground using \(\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{15} = 1\). Thus, \(\alpha = \tan^{-1}(1) = 45^\circ\).
3Step 3: Identify Elements and Build a Right Triangle for the Longer Wire
For the longer wire, another right triangle forms where the base is 15 feet and the height is 20 feet. This wire acts as the hypotenuse of its right triangle.
4Step 4: Calculate \(\sin \theta\) for Longer Wire
The sine of the angle \(\theta\) that the longer wire makes with the ground is calculated using \(\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\). First, calculate the hypotenuse: \(\text{hypotenuse} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\). Thus, \(\sin \theta = \frac{20}{25} = \frac{4}{5}\).
5Step 5: Calculate \(\cos \theta\) for Longer Wire
Using the same hypotenuse, calculate \(\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{25} = \frac{3}{5}\).
6Step 6: Determine the Cosine of the Angle between the Wires at the Ground
The cosine of the angle \(\phi\) between the wires at the ground is calculated using the dot product in vector form: \(\cos \phi = \frac{15 \times 15 + 15 \times 20}{\sqrt{15^2 + 15^2} \times \sqrt{15^2 + 20^2}}\). Calculate the dot product: \(225 + 300 = 525\). The magnitudes are \(\sqrt{450} = 15\sqrt{2}\) and \(25\). Thus, \(\cos \phi = \frac{525}{15\sqrt{2} \times 25} = \frac{525}{375\sqrt{2}} = \frac{14}{10\sqrt{2}} = \frac{7\sqrt{2}}{10}\).
7Step 7: Find the Angle between the Two Wires at the Ground
Use \(\cos^{-1}\) to find the angle \(\phi\) between the two wires, \(\phi = \cos^{-1}\left(\frac{7\sqrt{2}}{10}\right)\), which is approximately \(61^\circ\), when rounded to the nearest degree.
Key Concepts
Right TriangleTrigonometric RatiosAngle of ElevationVector Dot Product
Right Triangle
A right triangle is a triangle with one of its angles measuring 90 degrees. This forms a perfect "L" shape, where the side opposite to the right angle is called the hypotenuse, and the other two sides are known as the legs.
In many real-world problems, right triangles can be used to determine heights, distances, or angles when some dimensions are known.
In many real-world problems, right triangles can be used to determine heights, distances, or angles when some dimensions are known.
- The base of the triangle is the side along the horizontal surface.
- The height is vertical, often representing the distance to the object or height of the object.
- The hypotenuse is the wire or line that connects the two sides.
Trigonometric Ratios
Trigonometric ratios are functions used to relate the angles of a right triangle to its side lengths. The three primary trigonometric ratios are sine (
sin
), cosine (
cos
), and tangent (
tan
).
- Sine ( sin heta = rac{opposite}{hypotenuse} ): Measures the ratio of the side opposite to the angle ( ), relative to the hypotenuse.
- Cosine ( cos heta = rac{adjacent}{hypotenuse} ): Measures the ratio of the side adjacent to the angle ( ), relative to the hypotenuse.
- Tangent ( tan heta = rac{opposite}{adjacent} ): Is simply the ratio of the opposite side to the adjacent side.
Angle of Elevation
The angle of elevation is defined as the angle between the horizontal line and the line of sight, looking up at an object. Whenever we talk about calculation using trigonometric functions in real-world problems, the angle of elevation plays a significant role.
In the exercise concerning the telephone pole with bracing wires, the angle of elevation helps to determine how steeply the wires rise.
In the exercise concerning the telephone pole with bracing wires, the angle of elevation helps to determine how steeply the wires rise.
- In practice, angles of elevation are significant to engineers and architects, helping them determine how high a structure rises from a given base point.
- When computing this angle in our problem, we use the tangent function due to the simple relation involving the adjacent and opposite sides of the triangle.
Vector Dot Product
The vector dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the context of geometry and trigonometry, it is particularly useful for calculating the angles between vectors.
For the wires meeting at the ground, this concept determines the angle between them. Here's how it helps:
For the wires meeting at the ground, this concept determines the angle between them. Here's how it helps:
- The dot product is calculated by multiplying corresponding components and summing the results.
- The equation cos heta = rac{ ext{A} ullet ext{B}}{ ext{||A||} imes ext{||B||}} uses the magnitudes of vectors A and B along with their product.
- cos heta gives the cosine of the angle between vectors.
Other exercises in this chapter
Problem 21
In \(3-26,\) prove that each equation is an identity. $$ \frac{\tan ^{2} \theta}{\sec \theta-1}-1=\sec \theta $$
View solution Problem 21
An engineer wants to determine \(C D\) , the exact height of a building. To do this, he first locates \(B\) on \(\overline{C D}\) , a point 30 feet above \(C\)
View solution Problem 22
Prove the identity: \(\frac{\cos 2 \theta}{\sin \theta}+\frac{\sin 2 \theta}{\cos \theta}=\csc \theta\)
View solution Problem 22
If \(A=\arctan 2\) and \(B=\arctan (-2),\) find \(\tan (A-B)\)
View solution