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Chapter 8: Problem 107

Geometry: Angle between Two Lines Let \(L_{1}\) and \(L_{2}\) denote two nonvertical intersecting lines, and let \(\theta\) denote the acute angle between \(L_{1}\) and \(L_{2}\) (see the figure). Show that $$ \tan \theta=\frac{m_{2}-m_{1}}{1+m_{1} m_{2}} $$ where \(m_{1}\) and \(m_{2}\) are the slopes of \(L_{1}\) and \(L_{2},\) respectively. [Hint: Use the facts that \(\tan \theta_{1}=m_{1}\) and \(\left.\tan \theta_{2}=m_{2} .\right]\)

Short Answer

Expert verified
\(\tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2}\)

Step by step solution

01

Understand the problem

Given two intersecting lines with slopes \(m_1\) and \(m_2\), let's find the acute angle \(\theta\) between them by using the slopes.
02

Define angle between the lines

The acute angle \(\theta\) between the two lines with slopes \(m_1\) and \(m_2\) can be found using the tangent function. Given the hint, we know that \(\tan \theta_1 = m_1\) and \(\tan \theta_2 = m_2\).
03

Use the tangent subtraction formula

Using the tangent of the difference of two angles formula: \[\tan(\theta_2 - \theta_1) = \frac{\tan \theta_2 - \tan \theta_1}{1 + \tan \theta_1 \tan \theta_2}\]
04

Substitute the known values

Since \(\tan \theta_1 = m_1\) and \(\tan \theta_2 = m_2\), substitute these into the formula: \[\tan(\theta_2 - \theta_1) = \frac{m_2 - m_1}{1 + m_1 m_2}\]
05

Interpret the result

Given that the formula \(\tan(\theta_2 - \theta_1) = \frac{m_2 - m_1}{1 + m_1 m_2}\) accurately represents the tangent of the acute angle \(\theta\) between the lines as required by the problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometry
In geometry, understanding the relationships between different shapes, lines, and angles is crucial. When two lines intersect, they form an angle. The measure of this angle can give us crucial information about the spatial arrangement of these lines.
The intersection of two lines divides the plane into four angles. Only one of these angles will be the smallest 鈥� this is called the acute angle when it is less than 90 degrees. In this context, we are focused on calculating this smallest angle.
Slope
Slope is a measure of how steep a line is. It is calculated as the ratio of the rise (vertical change) to the run (horizontal change). Mathematically, for a line passing through points \(x_1, y_1\) and \(x_2, y_2\), the slope \m\ is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Each line in a coordinate system can be uniquely identified by its slope. When we know the slopes of two lines, we can find the angle between these lines using trigonometric functions. Remember, not just any angle, but specifically the acute angle if it exists.
Tangent Function
The tangent function relates an angle of a right triangle to the ratios of the triangle's sides. For an angle \theta\ in a right triangle, \tan \theta\ is the ratio of the opposite side to the adjacent side:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} \] When we apply this to the slopes of the lines, the tangent function helps us find the angle between them. If \m_1\ and \m_2\ are the slopes of two lines, the tangent of the acute angle \theta\ between them is determined by:
\[ \tan \theta = \frac{m_2 - m_1}{1 + m_1 m_2} \] This formula is derived using properties of the tangent function and illustrates how changes in slope directly translate to changes in angle.
Acute Angle
An acute angle is an angle that is less than 90 degrees. It is the smallest angle that can be formed when two lines intersect. Why is this important? Because it gives us a specific, meaningful measure of how two lines diverge from each other.
When calculating the angle between two lines using their slopes, we usually look for the acute angle because it provides a compact and easily understandable measure of the intersection.
By using the tangent function with the slopes of the lines, we can accurately identify this acute angle, giving us insight into the geometric relationship between the two lines.

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Most popular questions from this chapter

Problems 83 and 84 require the following discussion: When granular materials are allowed to fall freely, they form conical (cone-shaped) piles. The naturally occurring angle, measured from the horizontal, at which the loose material comes to rest is called the angle of repose and varies for different materials. The angle of repose \(\theta\) is related to the height \(h\) and the base radius \(r\) of the conical pile by the equation \(\theta=\cot ^{-1} \frac{r}{h} .\) See the illustration. Angle of Repose: De-icing Salt Due to potential transportation issues (for example, frozen waterways), de-icing salt used by highway departments in the Midwest must be ordered early and stored for future use. When de-icing salt is stored in a pile 14 feet high, the diameter of the base of the pile is 45 feet. (a) Find the angle of repose for de-icing salt. (b) What is the base diameter of a pile that is 17 feet high? (c) What is the height of a pile that has a base diameter of approximately 122 feet?

Find the exact value of each expression. $$ \cot \left[\sin ^{-1}\left(-\frac{1}{2}\right)\right] $$

Angle of Repose: Bunker Sand The steepness of sand bunkers on a golf course is affected by the angle of repose of the sand (a larger angle of repose allows for steeper bunkers). A freestanding pile of loose sand from a United States Golf Association (USGA) bunker had a height of 4 feet and a base diameter of approximately 6.68 feet. (a) Find the angle of repose for USGA bunker sand. Artillery A projectile fired into the first quadrant from the origin of a rectangular coordinate system will pass through the point \((x, y)\) at time \(t\) according to the relationship \(\cot \theta=\frac{2 x}{2 y+g t^{2}},\) where \(\theta=\) the angle of elevation of the launcher and \(g=\) the acceleration due to gravity \(=32.2\) feet/second \(^{2}\). An artilleryman is firing at an enemy bunker located 2450 feet up the side of a hill that is 6175 feet away. He fires a round, and exactly 2.27 seconds later he scores a direct hit. (a) What angle of elevation did he use? (b) If the angle of elevation is also given by \(\sec \theta=\frac{v_{0} t}{x}\) where \(v_{0}\) is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.

Solve each equation on the interval \(0 \leq \theta<2 \pi\) \((\tan \theta-1)(\sec \theta-1)=0\)

Calculus Show that the difference quotient for \(f(x)=\sin x\) is given by $$ \begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{\sin (x+h)-\sin x}{h} \\ &=\cos x \cdot \frac{\sin h}{h}-\sin x \cdot \frac{1-\cos h}{h} \end{aligned} $$
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