Question

Viewing angles An auditorium with a flat floor has a large flat panel television on one wall. The lower edge of the television is $3\mathrm{ft}$ above the floor, and the upper edge is $10\mathrm{ft}$ above the floor (see figure). Express $\theta$ in terms of $x.$

Short answer

Question: Express the angle θ in terms of the distance x in an auditorium with a flat floor consisting of a television fixed on a wall, with its lower and upper edges 3 meters and 10 meters above the floor, respectively. Answer: θ = tan^(-1)((13x)/(x^2-30))

Step-by-step solution

Identifying the Triangles

Observe the diagram given in the exercise. We can see two right-angled triangles, Triangle A (which has the angle θ1) and Triangle B (which has the angle θ2).

Calculating $\tan ({\theta }_1)$ and $\tan ({\theta }_2)$ from the Triangles

We know that $\tan (\theta )=\frac{\mathrm{opposite}}{\mathrm{adjacent}}$. Using this formula, find the tangent for each angle ${\theta }_1$ and ${\theta }_2$. For Triangle A, $\tan ({\theta }_1)=\frac{3}{x}$ For Triangle B, $\tan ({\theta }_2)=\frac{10}{x}$

Calculating θ

By observation, we can see that the sum of θ1 and θ2 is equal to θ: $\theta ={\theta }_1+{\theta }_2$ Now, we need to express θ in terms of x. To do this, we will use the formula $\tan (\theta )=\frac{\tan ({\theta }_1)+\tan ({\theta }_2)}{1-\tan ({\theta }_1)\cdot \tan ({\theta }_2)}$. So, $\tan (\theta )=\frac{\tan ({\theta }_1)+\tan ({\theta }_2)}{1-\tan ({\theta }_1)\cdot \tan ({\theta }_2)}$ $\tan (\theta )=\frac{\frac{3}{x}+\frac{10}{x}}{1-\frac{3}{x}\cdot \frac{10}{x}}$

Simplifying the Expression

Let's simplify the equation: $\tan (\theta )=\frac{\frac{3+10}{x}}{1-\frac{30}{x^2}}$ $\tan (\theta )=\frac{13}{x-\frac{30}{x}}$ Multiplying both the numerator and denominator by x: $\tan (\theta )=\frac{13x}{x^2-30}$ Since the exercise asks us to find the value of θ in terms of x, we can use the inverse tangent function: $\theta ={\tan }^{-1}\left(\frac{13x}{x^2-30}\right)$

Final Answer

Therefore, the angle θ can be expressed in terms of x as: $\boxed{{\displaystyle \theta ={\tan }^{-1}\left(\frac{13x}{x^2-30}\right)}}$

Key concepts

Right-Angled Triangles

Right-angled triangles are the bread and butter of trigonometry, a fascinating realm of mathematics that deals with the angles and sides of triangles. The right angle, which is exactly 90 degrees, is the cornerstone that makes these triangles so special and useful in calculations. Each right-angled triangle has two other angles that are less than 90 degrees, and these smaller angles are what we often need to find or use to calculate distances and other associated measures.

In the context of viewing angles, like in the auditorium example mentioned, we use the attributes of right-angled triangles to express unseen angles in terms of known lengths. Visualize the scenario: you have a television on a wall and you're trying to find the angle at which you're viewing it from a certain point on the floor. The triangle that forms between you, the bottom edge of the television, and the top edge is a right-angled triangle. In essence, understanding these triangles empowers you to tackle real-world problems, measuring heights and distances without physically measuring them, just by using the power of trigonometry.

Tangent Function

Among the swath of various trigonometric functions, the tangent function emerges as a particularly handy tool. It relates an angle of a right-angled triangle to the ratio of the length of the opposite side over the length of the adjacent side. Mathematically, it's represented as $\tan (\theta )=\frac{\text{opposite}}{\text{adjacent}}$.

For our exercise, where the flat panel television evokes the curiosity about the angles involved, $\tan ({\theta }_1)=\frac{3}{x}$ and $\tan ({\theta }_2)=\frac{10}{x}$ are not just equations, they are stories of how the vertical heights of 3ft and 10ft play with the horizontal distance from the viewer to the screen, 'x'. When you're calculating the tangent of these angles, you're dissecting the triangle into parts you can understand and manipulate, thus simplifying the layered complexities of physical space into a neat, mathematical package.

Inverse Trigonometric Functions

We've dissected the right-angled triangle, employed the tangent function to establish a base-level understanding, and now it's time to flip the script with inverse trigonometric functions. These functions are essential when you know the sides of the triangle but need to find the angle–and that's exactly what we needed for our auditorium problem.

The inverse tangent function, denoted as ${\tan }^{-1}$ or arctan, allows us to work backward from the tangent ratio to the angle itself. Essentially, it answers the question: 'What angle would give me this particular tangent value?' For the viewing angles of the television problem, we used this function to express $\theta$ in terms of $x$, ultimately leading us to our final expression $\theta ={\tan }^{-1}\left(\frac{13x}{x^2-30}\right)$.

Understanding inverse functions is a leap into a world where the depths of angles are fathomable, and the heights of practical application are endless. Rather than guessing and checking, students now have the trigonometric expressway to reach their solutions with speed and precision.