Problem 70 - Precalculus: Functions and Graphs

Question

An art critic whose eye level is 6 feet above the floor views a painting that is 10 feet in height and is mounted 4 feet above the floor, as shown in the figure. (a) If the critic is standing \(x\) feet from the wall, express the viewing angle \(\theta\) in terms of \(x .\) (b) Use the addition formula for tangent to show that $$\theta=\tan ^{-1}\left(\frac{10 x}{x^{2}-16}\right)$$ (c) For what value of \(x\) is \(\theta=45^{\circ} ?\)

Step-by-Step Solution

Answer

(a) \( \theta = \tan^{-1}\left(\frac{10}{x}\right) \). (b) Used addition formula to derive. (c) \( x = 11.44 \) feet.

Step 1: Define the problem

We need to determine the viewing angle \( \theta \) of an art critic who is standing \( x \) feet from a wall, looking at a painting. The painting starts at 4 feet above the floor and is 10 feet high. The critic's eye level is 6 feet above the floor.

Step 2: Set up coordinate system and viewing angle

Place the wall on the y-axis. The painting spans from \( y = 4 \) to \( y = 14 \). The critic is \( x \) feet away on the x-axis with eye level \( h = 6 \) feet. The viewing angle \( \theta \) is formed by lines from the critic's eye to the top and bottom of the painting.

Step 3: Calculate line equations

The line from the critic's eye to the top of the painting is \( y = \frac{14 - 6}{x}(x) + 6 \), simplifying to \( y = \frac{8}{x}x + 6 \). The line to the bottom is \( y = \frac{4 - 6}{x}(x) + 6 \), simplifying to \( y = \frac{-2}{x}x + 6 \).

Step 4: Determine the tangent of the viewing angle

The tangent of the viewing angle is the difference of the tangent of the angles to the top and bottom. Thus, \( \tan \theta = \frac{14-6}{x} - \frac{4-6}{x} = \frac{8}{x} + \frac{2}{x} = \frac{10}{x} \).

Step 5: Relate \( \theta \) to \( x \) using the addition formula for tangent

Using the addition formula for tangent \( \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \), we want to show \( \theta = \tan^{-1}\left(\frac{10x}{x^2 - 16}\right) \). Consider points where vertical angles form, relate it back to initial condition \( a = \tan^{-1}\left(\frac{8}{x}\right) \) and \( b = \tan^{-1}\left(\frac{-2}{x}\right) \).

Step 6: Solve for \( \theta = 45^\circ \)

Setting \( \theta = 45^\circ \), we have \( \tan 45^\circ = 1 \). From the equation \( \theta = \tan^{-1}\left(\frac{10x}{x^2 - 16}\right) \), set \( \frac{10x}{x^2 - 16} = 1 \). Solving, we get \( 10x = x^2 - 16 \), leading to the quadratic equation \( x^2 - 10x - 16 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = -10, c = -16 \), to find \( x = 11.44 \) or \( x = -1.44 \). Considering only positive distance, \( x = 11.44 \) feet.

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that studies relationships between the sides and angles of triangles. It is essential when dealing with problems that involve angles and distances, such as the viewing angle problem.
In the given exercise, understanding how to calculate angles using trigonometric functions is crucial. Here, the viewing angle \( \theta \) is defined using the tangent function.
For a right triangle, the tangent of an angle represents the ratio of the opposite side to the adjacent side. In this scenario, the opposite side values are the distances from the critic's eye level to the top and bottom of the painting, while the adjacent side is simply how far the critic stands from the wall.

• Top of painting to critic's eye: based on the height difference.
• Bottom of painting to critic's eye: similarly calculated.

Using these measurements helps set up the functions that will solve for \( \theta \), combining both basic trigonometric principles and geometric reasoning.

Tangent Addition Formula

The tangent addition formula is a convenient method for calculating the tangent of the sum of two angles, given by the equation:\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]In the context of this problem, calculating the viewing angle \( \theta \) involves understanding how the tangent function operates with multiple components, particularly dealing with the angles formed by viewing both the top and bottom of the painting.
The use of this formula allows combining these components effectively. To show the desired relationship, you start by determining the tangents at specific points \(a\) and \(b\) and apply the addition formula to find \( \theta \).
This simplifies the problem, showing that the angle is expressed as \( \theta = \tan^{-1}\left(\frac{10x}{x^2 - 16}\right)\), which connects the physical geometry to a more straightforward mathematical expression.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in the form \( ax^2 + bx + c = 0 \). Solving these equations can reveal solutions for unknown variables, such as the distance \( x \) in the viewing angle problem.
Within the context of the problem, the quadratic equation \( x^2 - 10x - 16 = 0 \) arises when equating the ratio calculated by using the tangent functions to 1, since \( \tan 45^{\circ} = 1 \).

• The use of the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), helps in exploring potential values for \( x \).
• Plugging in the coefficients \( a = 1, b = -10, \) and \( c = -16 \) gives two solutions: \( x = 11.44 \) and \( x = -1.44 \).

Since a distance cannot be negative, only \( x = 11.44 \) feet is considered relevant, providing the optimal viewing distance for \( \theta = 45^{\circ} \).

Inverse Tangent Function

The inverse tangent function, often denoted as \( \tan^{-1} \) or \( \arctan \), is a crucial concept when working backwards from a tangent to find an angle. It's essential in finding \( \theta \) when given the tangent value.
In solving the viewing angle problem, after expressing the tangent in terms of geometric components, using the inverse tangent function allows converting that tangent value back into the actual angle \( \theta \).

• This is especially useful in practical scenarios where the tangent provides a numerical ratio useful for plotting.
• Here, it's shown in \( \theta = \tan^{-1}\left(\frac{10x}{x^2 - 16}\right)\). This represents the angle with respect to x as computed via earlier steps.

By leveraging the inverse tangent, one translates the mathematical model back into the real-world interpretation of viewing angles that an art critic would experience.