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Intercepts.

The intercepts of a line are the points where the graph crosses the axes.

Look Closer.

Look at the intercepts on the graph above. Because the \(y\)-intercept of a graph lies on the \(y\)-axis, its \(x\)-coordinate must be zero. And because the \(x\)-intercept lies on the \(x\)-axis, its \(y\)-coordinate must be zero.

Example 3.1 .

Use algebra to find the intercepts of the graph of \(~y=\dfrac{1}{2}x+4\text{.}\)

Solution.

To find the intercepts of a graph.

To find the \(x\)-intercept of a graph:

Substitute \(0\) for \(y\) in the equation and solve for \(x\text{.}\)

To find the \(y\)-intercept of a graph:

Substitute \(0\) for \(x\) in the equation and solve for \(y\text{.}\)

1.

What are the intercepts of a line?

Answer.

2.

How do we find the \(x\)-intercept of the graph of an equation?

Example 3.2 .

Graph the equation \(3x+2y=7\) by the intercept method.

Solution.

To Graph a Linear Equation Using the Intercept Method.

1. Find the \(x\)- and \(y\)-intercepts of the graph.

2. Draw the line through the two intercepts.

3. Find a third point on the graph as a check. (Choose any convenient value for \(x\) and solve for \(y\text{.}\))

3.

How do we graph a line by the intercept method?

4.

Make a table that you can use with the intercept method.

Answer.

5.

What is the \(y\)-intercept of the line \(3x+2y=7\text{?}\)

Answer.

\(\left(0,3\dfrac{1}{2}\right)\)

Example 3.3 .

The temperature in Nome was \(-12 \degree\) at noon and has been rising at a rate of \(2 \degree\) per hour all day.

Solution.

1. An equation for the temperature is

\begin{equation*} T=-12+2h \end{equation*}

A graph of the equation is shown below.

2. To find the \(T\)-intercept, we set \(h=0\) and solve for\(T\text{.}\)

\begin{equation*} T=-12 + 2(\alert{0})=-12 \end{equation*}

The \(T\)-intercept is \((0,-12)\text{.}\) This point tells us that when \(h=0,~ T=-12\text{,}\) or the temperature at noon was \(-12 \degree\text{.}\)

To find the \(h\)-intercept, we set \(T=0\) and solve for \(h\text{.}\)

\begin{align*} \alert{0} \amp = -12+2h \amp \amp \blert{\text{Add 12 to both sides.}}\\ 12 \amp = 2h \amp \amp \blert{\text{Divide both sides by 2.}}\\ 6 \amp = h \end{align*}

The \(h\)-intercept is the point \((6,0)\text{.}\) This point tells us that when \(h=6,~T=0\text{,}\) or the temperature will reach zero degrees at six hours after noon, or 6 pm.

6.

The intercepts of a graph often represent the or values for a particular variable.

7.

What does the \(h\)-intercept tell us in Example 3.3?

\begin{gather*} 5x-8=30 ~~~~~~~~ \dfrac{x}{5}-30=8 ~~~~~~~~ \dfrac{x-30}{5}=8\\ 5x+8=30 ~~~~~~~~ \dfrac{x}{5}+30=8 ~~~~~~~~ \dfrac{x+30}{5}=8 \end{gather*}

1.

2.

1.

Around 1950, people began cutting down the world's rain forests to clear land for agriculture. In 1970, there were about 9.8 million square kilometers of rain forest left, and by 1990 that figure had been reduced to 8.2 million square kilometers. These two data points are shown on the graph below. The horizontal axis displays the year, \(x\text{,}\) and the vertical axis shows the amount of rainforest remaining, \(y\) (in millions of square kilometers).

1. If we continue to clear the rainforests at the same rate, the graph will be a straight line. Draw a straight line through the two points in the figure above. Make your line long enough to cross both axes.

2. Estimate the coordinates of the \(x\)- and \(y\)-intercepts of your graph.

\begin{align*} \amp x\text{-intercept:}~~ \underline{\hspace{4.545454545454546em}}\\ \amp y\text{-intercept:}~~ \underline{\hspace{4.545454545454546em}} \end{align*}

Use the intercepts to answer the following questions:

3. How many million square kilometers of rainforest were present initially?

4. If we continue to clear the rainforest at the same rate, when will it be completely demolished?

2.

Objectives.

In this Lesson we practiced the following skills:

Finding the intercepts of a grpah

Graphing a line by the intercept method

Interpreting the intercepts in context

Questions.

1. In Activity 1, Problem 1, Delbert says that the intercepts are \((4,-6)\text{.}\) What is wrong with his statement?

2. In Activity 2, Problem 1, how did you find the answer to part (c)?

3. In Activity 2, Problem 2, what intervals did you use to scale each axis?

19.

During spring break, Francine took the train to San Francisco and then bicycled home. The graph below left shows Francine's distance \(d\) from home, in miles, after cycling for \(h\) hours. Use the graph to answer the questions.

1. How far is it from San Francisco to Francine's home?

2. How many hours did Francine cycle to get home?

3. After cycling for 12 hours, how far is Francine from home?

4. How far did Francine cycle in the first four hours?

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Intercepts. The intercepts of a line are the points where the graph crosses the axes. Look Closer. Look at the intercepts on the graph above. Because the \(y\)-intercept of a graph lies on the \(y\)-axis, its \(x\)-coordinate must be zero. And because the \(x\)-intercept lies on the \(x\)-axis, its \(y\)-coordinate must be zero.

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Active Prelude to Calculus

Matthew Boelkins

Department of Mathematics
Grand Valley State University
boelkinm@gvsu.edu

Production Editor

Mitchel T. Keller

Department of Mathematics
University of Wisconsin-Madison
mitch@rellek.net

August 26, 2022

Active Prelude to Calculus

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Note 3.21 .

Reducing a new problem to an earlier one is a frequently used technique in mathematics.

Law of Sines.

If the angles of a triangle are \(A, B\text{,}\) and \(C\text{,}\) and the opposite sides are respectively \(a, b,\) and \(c\text{,}\) then

\begin{equation*} \blert{\dfrac {\sin A}{a} = \dfrac {\sin B}{b} = \dfrac {\sin C}{c}} \end{equation*}

\begin{equation*} \blert{\dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {c}{\sin C}} \end{equation*}

Two observers onshore sight a ship at an unknown distance from the shore. The observers are 400 yards apart at points \(A\) and \(B\text{,}\) and they each measure the angle from the shoreline to the ship, as shown below. How far is the ship from the observer at \(A\text{?}\)

Solution .

First note that \(\triangle ABC\) is not a right triangle, so we cannot use the trig ratios directly to find the sides of this triangle.

The unknown distance \(d\) is the side opposite \(\angle B = 79.4\degree\text{.}\) In order to use the Law of Sines, we must know another angle and the side opposite that angle. We do: we know that side \(c = 400\text{,}\) and we can compute the angle at the ship, \(\angle C\text{.}\)

\begin{equation*} \angle C = 180\degree - (79.4\degree + 83.2\degree) = 17.4\degree \end{equation*}

Now we apply the Law of Sines, using angles \(B\) and \(C\text{.}\)

\begin{align*} \dfrac{b}{\sin B} \amp = \dfrac{c}{\sin C} \amp\amp \blert{\text{Substitute the given values.}}\\ \dfrac{d}{\sin 79.4\degree} \amp = \dfrac{400}{\sin 17.4\degree} \amp\amp \blert{\text{Evaluate sines.}}\\ \dfrac{d}{0.9829} \amp = \dfrac{400}{0.2990} \amp\amp \blert{\text{Multiply both sides by 0.9829.}}\\ d \amp = 1315 \end{align*}

The ship is about 1315 yards from the observer at \(A\text{.}\)

Delbert and Francine are 40 feet apart on one side of a river. They make angle measurements to a pine tree on the opposite shore as shown at right. What is the distance from Francine to the pine tree?

In the triangle shown at right, \(A = 37\degree, B = 54\degree\text{,}\) and \(a = 11\text{.}\)

Solution .

1. We use the Law of Sines with \(a\) and angle \(A\) to find \(b\text{.}\)

\begin{align*} \dfrac{a}{\sin A} \amp = \dfrac{b}{\sin B} \amp\amp \blert{\text{Substitute the given values.}}\\ \dfrac{11}{\sin 37\degree} \amp = \dfrac{b}{\sin 54\degree} \amp\amp \blert{\text{Multiply both sides by} \sin 54\degree.}\\ \dfrac{11}{\sin 37\degree} \cdot \sin 54\degree\amp = b \end{align*}

Evaluating this expression with a calculator, we find that \(b \approx 14.79\text{.}\)

2. The angle \(C = 180\degree - (37\degree + 54\degree) = 89\degree\text{.}\) Now we use the Law of Sines to find side \(c\text{.}\) Note that it is safer to use the given side, \(a\text{,}\) rather than the value we calculated for \(b\text{.}\)

\begin{align*} \dfrac{a}{\sin A} \amp = \dfrac{c}{\sin C} \amp\amp \blert{\text{Substitute the known values.}}\\ \dfrac{11}{\sin 37\degree} \amp = \dfrac{c}{\sin 89\degree} \amp\amp \blert{\text{Multiply both sides by} \sin 89\degree.}\\ \dfrac{11}{\sin 37\degree} \cdot \sin 89\degree\amp = c \end{align*}

In the triangle at right, \(A = 65\degree, C = 42\degree\text{,}\) and \(c = 16\text{.}\) Solve the triangle. (Hint: Which side will you find first?)

In the triangle shown at right, \(B = 55\degree, a = 5\text{,}\) and \(b = 11\text{.}\) Solve the triangle.

Solution .

We must find the three remaining parts of the triangle, \(A, C\text{,}\) and \(c\text{.}\) First, we use the Law of Sines to find \(\sin A\text{.}\)

\begin{align*} \dfrac{\sin A}{a} \amp = \dfrac{\sin B}{b} \amp\amp \blert{\text{Substitute the known values.}}\\ \dfrac{\sin A}{5} \amp = \dfrac{\sin 55\degree}{11} \amp\amp \blert{\text{Multiply both sides by 5.}}\\ \sin A \amp = 5 \cdot \dfrac{\sin 55\degree}{11} \approx 0.3723 \end{align*}

So \(A = \sin^{-1}(0.3723) \approx 21.9\degree\text{.}\) Now we know two angles, and we can find angle \(C\text{.}\)

\begin{align*} C \amp = 180\degree - (B + A)\\ \amp = 180\degree - 55\degree - 21.9\degree = 103.1\degree \end{align*}

Finally, we use the Law of Sines to find side \(c\text{.}\)

\begin{align*} \dfrac{b}{\sin B} \amp = \dfrac{c}{\sin C} \amp\amp \blert{\text{Substitute the known values.}}\\ \dfrac{11}{\sin 55\degree} \amp = \dfrac{c}{\sin 103.1\degree} \amp\amp \blert{\text{Multiply both sides by} \sin 103.1\degree.}\\ \dfrac{11}{\sin 55\degree} \cdot \sin 103.1\degree\amp = c \end{align*}

Evaluating this expression with a calculator gives \(c \approx 13.1.\)

Sketch a triangle with \(C = 93\degree, a = 7\text{,}\) and \(c = 11\text{.}\)

1. Use the Law of Sines to find another angle of the triangle.

2. Solve the triangle, and label your sketch with the results.

Answer .

1. \(\displaystyle A = 39.5\degree\)

2. \(\displaystyle b = 8.13,~B = 47.5\degree\)

In the last example we used the value of \(\sin A\) to find angle \(A\text{.}\) But there are two angles that satisfy \(\sin A = 0.3723\text{,}\) one acute (\(21.9\degree\)) and one obtuse (\(168.1\degree\)). How did we know which one to choose?

In this case, the obtuse angle is too big to fit in the triangle, because angle \(B = 55\degree\text{.}\) Sometimes both angles will produce (different) triangles, and sometimes only the acute angle will work. You should always check whether both angles provide solutions.

Because there can be more than one solution for a triangle in which we know two sides and an angle opposite one of them, this situation is called the ambiguous case for the Law of Sines. (See Homework Problems 33-42 for more about the ambiguous case.)

Find two triangles in which \(B = 14.4\degree, a = 8\text{,}\) and \(b = 3\text{,}\) and sketch both triangles.

Solution .

Using the Law of Sines, we have

\begin{align*} \dfrac{\sin A}{a} \amp = \dfrac{\sin B}{b} \amp\amp \blert{\text{Substitute the known values.}}\\ \dfrac{\sin A}{8} \amp = \dfrac{\sin 14.4\degree}{3} \amp\amp \blert{\text{Multiply both sides by 8.}}\\ \sin A \amp = 8 \cdot \dfrac{\sin 14.4\degree}{3} \approx 0.6632 \end{align*}

There are two angles with sine \(0.6632\text{:}\)

the acute angle \(A = \sin^{-1}(0.6632) = 41.5\degree,\)

or its supplement, \(A^{\prime} = 180\degree - 41.5\degree = 138.5\degree\text{.}\)

Each of these angles produces a different solution triangle, because the angle \(C\) and side \(c\) will be different also.

In the first case, angle \(C = 180\degree - (14.4\degree + 138.5\degree) = 27.1\degree\text{,}\) and we have the triangle shown in figure (a).

In the second case, \(C = 180\degree - (14.4\degree + 41.5\degree) = 124.1\degree\text{,}\) which gives us the triangle shown in figure (b).

Suppose that \(C = 29.7\degree, b = 8\text{,}\) and \(c = 5\text{.}\)

1. Find two possible values for angle \(B\text{.}\)

2. Solve the triangle for both values of \(B\text{,}\) and sketch both solutions.

Answer .

\(A = 97.9\degree,~a = 10\)

or \(A = 22.7\degree,~a = 3.9\)

Richard wants to measure the height of a castle controlled by hostile forces. When he is as close as he can get to the castle, the angle of elevation to the top of the wall is \(18.5\degree\text{.}\) He then retreats 20 yards and measures the angle of elevation again; this time it is \(15.9\degree\text{.}\) How tall is the castle?

Solution .

Notice that \(h\) is one side of the right triangle \(ADC\text{.}\) If we can find its hypotenuse, labeled \(r\) in the figure, we can use the sine ratio to find \(h\text{.}\) To find \(r\text{,}\) we consider a second triangle, \(ABC\text{,}\) as shown below.

Now, the angle opposite \(r, ~\angle ABC\text{,}\) is the complement of \(18.5\degree\text{,}\) so

\begin{equation*} \angle ABC = 180\degree - 18.5\degree = 161.5\degree \end{equation*}

The angle opposite the 20-yard side, \(\angle BAC\text{,}\) is

\begin{equation*} \angle BAC = 180\degree - (161.5\degree + 15.9\degree) = 2.6\degree \end{equation*}

Now we can apply the Law of Sines to find \(r\text{.}\) We have

\begin{align*} \dfrac{r}{\sin 161.5\degree} \amp = \dfrac{20}{\sin 2.6\degree} \amp\amp \blert{\text{Solve for }r.}\\ r \amp = \sin 161.5\degree \cdot \dfrac{20}{\sin 2.6\degree}\\ \amp \approx 139.9 \end{align*}

So \(r\) is about 139.9 yards. Finally, using the right triangle \(ADC\) and the definition of sine, we can write

\begin{align*} \dfrac{h}{r} \amp = \sin 15.9\degree \amp\amp \blert{\text{Solve for }h.}\\ h \amp = r \cdot \sin 15.9\degree \approx 38.33 \end{align*}

The castle is about 38.33 yards tall.

Solve the problem in the previous example again, but instead of finding \(r\text{,}\) find the length \(AB\text{,}\) and then use \(\triangle ABD\) to find \(h\text{.}\)

Answer .

\(AB = 120.79\text{,}\) the castle is about 38.33 yards tall.

Astronomers 1000 kilometers apart observe an asteroid with a parallax of \(0.001\degree\text{.}\) How far is the asteroid from Earth?

Solution .

We let \(x\) represent the distance to the asteroid. The asteroid and the two observers make an isosceles triangle with base 1000 km and equal sides of length \(x\text{,}\) as shown below.

The base angles of the triangle are both \(\dfrac{180\degree - 0.001\degree}{2} = 89.999\degree\text{.}\) By the Law of Sines,

\begin{align*} \dfrac{x}{\sin 89.999\degree} \amp = \dfrac{1000}{\sin 0.001\degree} \amp\amp \blert{\text{Solve for }x.}\\ x \amp = \sin 89.999\degree \cdot \dfrac{1000}{\sin 0.001\degree}\\ x \amp \approx 57,000,000 \end{align*}

The asteroid is about 57 million kilometers from Earth (roughly one third of the distance to the Sun).

Two observers 800 kilometers apart observe an object with a parallax of \(0.0005\degree\text{.}\) How far is the object from Earth?

Fractions of a Degree.

One minute: \(~~~~~~1^{\prime} = \dfrac{1\degree}{60}\)

One second: \(~~~~~~1^{\prime \prime} = \dfrac{1^{\prime}}{60} = \dfrac{1\degree}{3600}\)

The star Wolf 359 has a parallax of \(0.85^{\prime \prime}\) when observed from opposites sides of Earth's orbit. How far away is the star?

Solution .

In the figure below, the star Wolf 359 is located at point \(O\text{.}\) Our Sun is located at point \(C\text{,}\) halfway between the two observation points at \(A\) and \(B\text{.}\) Thus, the distance between the observation points is twice the distance from the Earth to the Sun, or 2 AU.

The altitude from \(O\) to side \(\overline{AB}\) forms a right triangle \(ACO\) and bisects the angle at \(O\text{.}\) Thus

\begin{equation*} \angle AOC = \dfrac{1}{2}(0.85^{\prime \prime}) = 0.425^{\prime \prime} \end{equation*}

We use the definition of tangent to find

\begin{align*} \tan 0.425^{\prime \prime} \amp = \dfrac{\text{opposite}}{\text{adjacent}}\\ \tan (0.425\degree / 3600) \amp = \dfrac{1}{x} \amp\amp \blert{\text{Solve for }x.}\\ x \amp = \dfrac{1}{\tan 0.425\degree / 3600} \approx 485,000 \end{align*}

The star is approximately \(485,000\) AU from Earth, or nearly half a million times the distance from Earth to the Sun.

Two observers are 1 AU (astronomical unit) apart. They find that the parallax to a distant star is \(1^{\prime \prime}\text{.}\) What is the distance to the star, in astronomical units?

(This distance is called a parsec. In other words, a parsec is the distance at which the parallax from observations 1 AU apart is \(1^{\prime \prime}\text{.}\) In this Exercise, you are calculating the number of astronomical units in 1 parsec.)

Answer .

1 parsec \(\approx 206,265\) AU

Convert to a decimal fraction.

1. \(4^{\prime} =\) \(\degree\)

2. \(2^{\prime \prime} =\) \(\degree\)

3. \(1^{\prime} ~5^{\prime \prime}=\) \(\degree\)

\(\underline{\qquad\qquad\qquad\qquad}\)

Algebra Refresher Answers

1. \(\displaystyle 0.41\overline{6}\)

2. \(\displaystyle 0.58\overline{3}\)

3. \(\displaystyle 0.2\overline{6}\)

4. \(\displaystyle 0.0\overline{6}\)

5. \(\displaystyle 0.000\overline{5}\)

6. \(\displaystyle 0.0180\overline{5}\)

Law of Sines.

If the angles of a triangle are \(A, B\text{,}\) and \(C\text{,}\) and the opposite sides are respectively \(a, b,\) and \(c\text{,}\) then

\begin{equation*} \dfrac {\sin A}{a} = \dfrac {\sin B}{b} = \dfrac {\sin C}{c} \end{equation*}

\begin{equation*} \dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {c}{\sin C} \end{equation*}

Fractions of a Degree.

One minute: \(~~~~~~1^{\prime} = \dfrac{1\degree}{60}\)

One second: \(~~~~~~1^{\prime \prime} = \dfrac{1^{\prime}}{60} = \dfrac{1\degree}{3600}\)

23.

Chad is hiking along a straight path but needs to detour around a large pond. He turns \(23\degree\) from his path until clear of the pond, then walks back to his original path, intercepting it at an angle of \(29\degree\) and at a distance of 2 miles from where he had left the path. How far did Chad walk in each of the two segments of his detour, and how much farther did his detour require compared with a straight line through the pond?

24.

Bob is flying to Monterey but must change course to avoid a storm. He flies \(19\degree\) off from his original direction until he clears the storm, then turns again to return back to his original flight path, intercepting it at an angle of \(54.9\degree\) and at a distance of 50 miles from where he had left it. How much farther did his detour require compared with his original course?

25.

Geologists find an outcropping from an underground rock formation that normally indicates the presence of oil. The outcropping is on a hillside, and the formation itself dips another \(17\degree\) from the surface. If an oil well is placed 1000 meters downhill from the outcropping, how far will the well have to drill before it reaches the formation?

26.

A proposed ski lift will rise from point near the base of the slope with an angle of \(27\degree\text{.}\) At a distance of 400 meters further from the slope, the angle of elevation to the top of the ski lift is \(19\degree\text{.}\) How long is the ski lift?

27.

Thelma wants to measure the height of a hill. She first plants a 50 foot tall antenna at the hill's peak. Then she descends the hill and finds a point where she can see the top and the bottom of the antenna. The angle of elevation to the bottom of the antenna is \(23\degree\text{,}\) and the angle of elevation to the top of the antenna is \(24\degree\text{.}\)

1. Find \(\angle ACB\text{.}\)

2. Find\(\angle CAB\text{,}\) at the top of the antenna.

3. How tall is the hill?

28.

A billboard of California's gubernatorial candidate Angelyne is located on the roof of a building. At a distance of 180 feet from the building, the angles of elevation to the bottom and top of the billboard are respectively \(39.8\degree\) and \(47.3\degree\text{.}\) How tall is the billboard?

33.

1. Can you draw a triangle \(ABC\) with \(A = 30\degree\) and \(c = 3\) if \(a \lt \dfrac{3}{2}\text{?}\) Why or why not?

2. How many triangles are possible if \(\dfrac{3}{2} \lt a \lt 3\text{?}\)

3. How many triangles are possible if \(a \gt 3\text{?}\)

34.

In this problem we show that there are two different triangles \(ABC\) with \(A = 30\degree,~ a = 2\) and \(c = 3\text{.}\)

Use the Law of Sines to find two distinct possible measures for \(\angle C\text{.}\)

35.

In \(\triangle ABC, \angle A = 30\degree\) and \(c = 12\text{.}\) How many triangles are possible for each of the following lengths for side \(a\text{?}\) Sketch the solutions in each case.

36.

Consider the triangle \(ABC\) shown below.

1. Now suppose we keep \(\angle A\) and side \(c\) fixed, but allow \(a\) to vary in length. What is the smallest value \(a\) can have and still be long enough to make a triangle?

2. What are the largest and smallest values that \(a\) can have in order to produce two distinct triangles \(ABC\) (without changing \(\angle A\) and side \(c\))?

37.

For the triangle in Problem 36, suppose \(A = 40\degree\) and \(c = 8\text{.}\)

1. Sketch and solve the triangle if \(a = 12\text{.}\)

2. Sketch and solve the triangle if \(a = 6\text{.}\)

3. Sketch and solve the triangle if \(a = 4\text{.}\)

38.

For the figure in Problem 36, suppose \(A = 70\degree\) and \(c = 20\text{.}\)

1. For what value of \(a\) is the triangle a right triangle?

2. For what values of \(a\) are there two solutions for the triangle?

3. For what values of \(a\) is there one obtuse solution for the triangle?

4. For what value of \(a\) is there no solution?

43.

Delbert and Francine are 1000 yards apart. The angle Delbert sees between Francine and a certain tree is \(38\degree\text{.}\) If the tree is 800 yards from Francine, how far is it from Delbert? (There are two possible answers.)

44.

From the lookout point on Fabrick Rock, Ann can see not only see the famous "Crooked Spire" in Chesterfield, which is 8 miles away, but also the red phone box in the village of Alton. Chesterfield and Alton are 7 miles apart. Fabrick Rock has a plaque that shows directions to famous sites, and from the plaque Ann determines that the angle between the lines to the spire and the phone box measures \(19\degree\text{.}\) How far is Fabrick Rock from the phone box? (There are two possible answers.)

45.

1. Use the Law of Sines to find \(a\text{.}\)

2. Use the Law of Sines to find \(c\text{.}\)

3. Find \(c\) without using the Law of Sines. (Hint: Sketch the altitude, \(h\text{,}\) from \(C\) to make two right triangles. Find \(h\text{,}\) then use \(h\) to find \(c\text{.}\))

46.

1. Use the Law of Sines to find \(c\text{.}\)

2. Find \(c\) without using the Law of Sines.

47.

Sketch a triangle with angles \(A, ~B\) and \(C\) and opposite sides of lengths respectively \(a,~ b\) and \(c\text{.}\)

49.

Here is a method for solving certain oblique triangles by dividing them into two right triangles. In the triangle shown, we know two angles, \(A\) and \(B\text{,}\) and the side opposite one of them, say \(a\text{.}\) We would like to find side \(b\text{.}\)

Which of the following is equivalent to the formula you wrote in part (d)?

1. \(\displaystyle a \sin A = b \sin B\)

2. \(\displaystyle \dfrac{a}{\sin A} = \dfrac{b}{\sin B}\)

3. \(\displaystyle \dfrac{a}{\sin B} = \dfrac{b}{\sin A}\)

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Reducing a new problem to an earlier one is a frequently used technique in mathematics. If the angles of a triangle are \(A, B\text{,}\) and \(C\text{,}\) and the opposite sides are respectively \(a, b,\) and \(c\text{,}\) then \begin{equation*} \blert{\dfrac {\sin A}{a} = \dfrac {\sin B}{b} = \dfrac {\sin C}{c}} \end{equation*} \begin{equation*} \blert{\dfrac {a}{\sin A} = \dfrac {b}{\sin B} = \dfrac {c}{\sin C}} \end{equation*} Two observers onshore sight a ship at an unknown distance from the shore.

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ORCCAOpen Resources for Community College Algebra

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ORCCAOpen Resources for Community College Algebra

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Chapter 8 Sequences and Series of Discrete Mathematics: An Open Introduction

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Chapter 8 Sequences and Series of Discrete Mathematics: An Open Introduction

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The elements of the sequence or series are distinct and separate, typically representing values or observations at specific points in time or discrete intervals.

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Look Closer.

The method works well in this example because both equations in the system were already in the form \(y=mx+b\text{.}\) Sometimes we must solve one of the equations for \(y\) in terms of \(x\) before we can substitute.

To Solve a System by Substitution.

1. Choose one of the variables in one of the equations. (It is best to choose a variable whose coefficient is 1 or \(-1\text{.}\)) Solve the equation for that variable.

2. Substitute the result of Step 1 into the other equation. This gives an equation in one variable.

3. Solve the equation obtained in Step 2. This gives the solution value for one of the variables. Substitute this value into the result of Step 1 to find the solution value of the other variable.

Look Ahead.

You may have noticed that this method worked because the coefficients of \(y\) in the two equations were opposites, \(2\) and \(-2\text{.}\) This caused the \(y\)-terms to cancel out when we added the two equations together. What if the coefficients of neither \(x\) nor \(y\) are opposites? Then we must multiply one or both of the equations in the system by a suitable constant.

In Example 4.18, we multiplied both sides of the equation \(3x+y=-1\) by \(-3\text{.}\) (This is an application of the Multiplication Property of Equality.) We must be careful to multiply every term on both sides of the equation by the same constant. Otherwise, we won't have an equivalent equation -- its solutions will not be the same.

Linear Combination.

When we add a multiple of one equation to the other we are making a linear combination of the equations.

Use linear combinations to solve the system

\begin{gather*} 5x-2y=22\\ 2x-5y=13 \end{gather*}

Solution.

To Solve a System by Elimination.

1. Write each equation in the form \(Ax+By = C\text{.}\)

2. Decide which variable to eliminate. Multiply each equation by an appropriate constant so that the coefficients of that variable are opposites.

3. Add the equations from Step 2 and solve for the remaining variable.

4. Substitute the value found in Step 3 into one of the original equations and solve for the other variable.

Look Ahead.

How do you know which method to use to solve a system, substitution or elimination? Both methods will work on any linear system. However, substitution will be easier if one of the variables in one of the equations has a coefficient of \(1\) or \(-1\text{.}\) Otherwise, the elimination method is usually more efficient.

Solve each system by elimination.

1. \(\displaystyle \begin{aligned}[t] 3x-y=2\\ -6x+2y=3\end{aligned}\)

2. \(\displaystyle \begin{aligned} x-2y=3\\ 2x-4y=6\end{aligned}\)

Solution.

1. To eliminate the \(y\)-terms, we multiply the first equation by 2 and add:

\begin{align*} 6x-2y \amp = 4 \\ \underline{-6x+2y}\amp\underline{{}= 3\vphantom{y}}\\ 0x+0y \amp = 7 \end{align*}

Both variables are eliminated, and we are left with the false statement \(0=7\text{.}\) There are no values of \(x\) or \(y\) that will make this equation true, so the system has no solutions. The graph shows that the system is inconsistent.

2. To eliminate the \(x\)-terms, we multiply the first equation by \(-2\) and add:

\begin{align*} -2x+4y \amp = -6\\ \underline{2x-4y}\amp \underline{{}= \hphantom{-}6\vphantom{y}}\\ 0x+0y \amp = 0 \end{align*}

We are left with the true but unhelpful equation \(0=0\text{.}\) The two equations are in fact equivalent (one is a constant multiple of the other), so the system is dependent. The graph of both equations is shown in the figure.

When Using Elimination to Solve a System.

1. If combining the two equations results in an equation of the form

\begin{equation*} 0x+0y=k~~~~(k \not= 0) \end{equation*}

then the system is inconsistent.

2. If combining the two equations results in an equation of the form

\begin{equation*} 0x+0y=0 \end{equation*}

then the system is dependent.

7.

How can you tell if a system is inconsistent?

Answer.

If elimination leads to

8.

How can you tell if a system is dependent?

Answer.

If elimination leads to

2.

Follow the suggested steps to solve the system by substitution:

\begin{align*} 3y-2x \amp = 3\\ x-2y \amp = -2 \end{align*}

1. \(\displaystyle \blert{\text{Solve the second equation for}~x~\text{in terms of}~y.}\)

2. \(\displaystyle \blert{\text{Substitute your expression for}~x~\text{into the first equation.}}\)

3. \(\displaystyle \blert{\text{Solve the equation you got in Step 2.}}\)

4. \(\displaystyle \blert{\text{Substitute the}~y~\text{-value into your result from Step 1 to find}~x.}\)

5. \(\displaystyle \blert{\text{Verify that your solution values satisfy both equations in the system.}}\)

1.

Follow the suggested steps to solve the system by elimination:

\begin{align*} 2x-3y \amp = 6\\ 4x-5y \amp = 8 \end{align*}

For this problem, we will eliminate the \(x\)-terms, so we arrange for their coefficients to be opposites.

1. \(\displaystyle \blert{\text{Multiply each term of the first equation by}~-2.}\)

2. \(\displaystyle \blert{\text{Add the new equations and solve the result for}~y.}\)

3. \(\displaystyle \blert{\text{Substitute your value for}~y~\text{into the first equation and solve for}~x.}\)

4. \(\displaystyle \blert{\text{Verify that your solution values satisfy both equations in the system.}}\)

2.

Follow the suggested steps to solve the system by elimination:

\begin{align*} 3x \amp = 2y+13\\ 3y-15 \amp = -7x \end{align*}

For this problem, we will eliminate the \(y\)-terms.

1. \(\displaystyle \blert{\text{Write each equation in the form}~Ax+By=C.}\)

2. \(\displaystyle \blert{\text{Find rhe LCM of the}~y~\text{-coefficients. Multiply each equation by an appropriate constant.}}\)

3. \(\displaystyle \blert{\text{Add the new equations and solve the result for}~x.}\)

4. \(\displaystyle \blert{\text{Substitute your value for}~x~\text{into the second equation and solve for}~y.}\)

5. \(\displaystyle \blert{\text{Verify that your solution values satisfy both equations in the system.}}\)

3.

1. Solve the system by elimination:

\begin{align*} x+3y \amp = 6\\ 2x-12 \amp = -6y \end{align*}

2. Is the system dependent, inconsistent, or consistent and independent?

1.

A train ticket from Camarillo to San Diego costs $31 in coach and $47 for business class. On Tuesday, there were 42 passengers on the morning train, and Amtrak took in $1494 in fares. How many coach passengers took the morning train, and how many business class passengers?

1. Choose variables:

\begin{gather*} x = \\ y = \end{gather*}

2. Complete the table.

\(\hphantom{0000}\)

Number of passengers

Price per ticket

Total revenue

Coach

\(\hphantom{0000}\)

\(\hphantom{0000}\)

\(\hphantom{0000}\)

Business

\(\hphantom{0000}\)

\(\hphantom{0000}\)

\(\hphantom{0000}\)

3. Write an equation about the number of passengers:

\begin{equation*} \hphantom{0000} \end{equation*}

Write an equation about revenue:

\begin{equation*} \hphantom{0000} \end{equation*}

4. Solve your system.

Write your answer in a sentence.

2.

The perimeter of a rectangular playground is 197 yards, and its length is 5 yards less than twice its width. Find the dimensions of the playground.

1. Choose variables:

\begin{gather*} x = \\ y = \end{gather*}

2. Write an equation about the perimeter of the playground:

\begin{equation*} \hphantom{0000} \end{equation*}

Write an equation relating the length and the width:

\begin{equation*} \hphantom{0000} \end{equation*}

3. Solve your system.

Write your answer in a sentence.

Objectives.

In this Lesson we practiced the following skills:

Solving systems by the substitution method

Solving systems by the elimination method

Identifying a system as consistent, inconsistent, or dependent

Writing a system of equations to solve an applied problem

Questions.

1. In Activity 1, Problem 2, why did we choose to solve the second equation for \(x\) in terms of \(y\text{?}\)

2. In Activity 2, Problem 1, why did we multiply the first equation by \(-2\text{?}\)

3. In Activity 2, Problem 2, what was the LCM of the \(y\)-coefficients?

4. In Activity 3, Problem 1, what were the two equations about?

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Look Closer. The method works well in this example because both equations in the system were already in the form \(y=mx+b\text{.}\) Sometimes we must solve one of the equations for \(y\) in terms of \(x\) before we can substitute. To Solve a System by Substitution. Choose one of the variables in one of the equations.

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Discrete Mathematics: An Open Introduction

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Discrete Mathematics: An Open Introduction

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Problem 47 has a geometric interpretation in a coordinate plane. A lattice path in the plane is a "curve" made up of line segments that either go from a point \((i,j)\) to the point \((i+1,j)\) or from a point \((i,j)\) to the point \((i,j+1)\text{,}\) where \(i\) and \(j\) are integers. (Thus lattice paths always move either up or to the right.) The length of the path is the number of such line segments.

How many such lattice paths of that length are there?

In Problem 47 you saw that we had to make ten choices of north or east, choosing north four times.

Another kind of geometric path in the plane is a diagonal lattice path. Such a path is a path made up of line segments that go from a point \((i,j)\) to \((i+1,j+1)\) (this is often called an upstep) or \((i+1,j-1)\) (this is often called a downstep), again where \(i\) and \(j\) are integers. (Thus diagonal lattice paths always move towards the right but may move up or down.)

A school play requires a ten dollar donation per person; the donation goes into the student activity fund. Assume that each person who comes to the play pays with a ten dollar bill or a twenty dollar bill. The teacher who is collecting the money forgot to get change before the event. If there are always at least as many people who have paid with a ten as a twenty as they arrive the teacher won't have to give anyone an IOU for change. Suppose \(2n\) people come to the play, and exactly half of them pay with ten dollar bills.

Problem 51

Notice that a lattice path from \((0,0)\) to \((n,n)\) stays inside (or on the edges of) the square whose sides are the \(x\)-axis, the \(y\)-axis, the line \(x=n\) and the line \(y=n\text{.}\) In this problem we will compute the number of lattice paths from (0,0) to \((n,n)\) that stay inside (or on the edges of) the triangle whose sides are the \(x\)-axis, the line \(x=n\) and the line \(y=x\text{.}\) For example, in Figure 1.3.1 we show the grid of points with integer coordinates for the triangle whose sides are the \(x\)-axis, the line \(x=4\) and the line \(y=x\text{.}\)

Figure 1.3.1 The lattice paths from \((0,0)\) to \((i,i)\) for \(i=0,1,2,3,4\text{.}\) The number of paths to the point \((i,i)\) is shown just above that point.

(c)

Find a formula for the number of lattice paths from \((0,0)\) to \((n,n)\) that do not go above the line \(y=x\text{.}\) The number of such paths is called a Catalan Number and is usually denoted by \(C_n\text{.}\)

Find a formula for the number of lattice paths from \((0,0)\) to \((n,n)\) that do not go above the line \(y=x\text{.}\) The number of such paths is called a Catalan Number and is usually denoted by \(C_n\text{.}\)

Your formula for the Catalan Number can be expressed as a binomial coefficient divided by an integer. Whenever we have a formula that calls for division by an integer, an ideal combinatorial explanation of the formula is one that uses the quotient principle. The purpose of this problem is to find such an explanation using diagonal lattice paths. A diagonal lattice path that never goes below the \(y\)-coordinate of its first point is called a Dyck Path. We will call a Dyck Path from \((0,0)\) to \((2n,0)\) a Catalan Path of length \(2n\text{.}\) Thus the number of Catalan Paths of length \(2n\) is the Catalan Number \(C_n\text{.}\)

We know that \((x+y)^2 = x^2+2xy+y^2\text{.}\) Multiply both sides by \((x+y)\) to get a formula for \((x+y)^3\) and repeat to get a formula for \((x+y)^4\text{.}\) Do you see a pattern? If so, what is it? If not, repeat the process to get a formula for \((x+y)^5\) and look back at Figure 1.2.4 to see the pattern. Conjecture a formula for \((x+y)^n\text{.}\)

When we apply the distributive law \(n\) times to \((x+y)^n\text{,}\) we get a sum of terms of the form \(x^iy^{n-i}\) for various values of the integer \(i\text{.}\) If it is clear to you that each term of the form \(x^iy^{n-i}\) that we get comes from choosing an \(x\) from \(i\) of the \((x+y)\) factors and a \(y\) from the remaining \(n-i\) of the factors and multiplying these choices together, then answer part a of the problem and skip part b. In either case, be sure to answer part c.

We can take this step-by-step and consider a small case to get started.

Expand the product \((x_1 +y_1)(x_2 +y_2)(x_3+y_3)\text{.}\)

What do you get when you substitute \(x\) for each \(x_i\) and \(y\) for each \(y_i\text{?}\)

Now imagine expanding

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n). \end{equation*}

Once you apply the commutative law to the individual terms you get, you will have a sum of terms of the form

\begin{equation*} x_{k_1}x_{k_2}\cdots x_{k_i}\cdot y_{j_1}y_{j_2}\cdots y_{j_{n-i}}. \end{equation*}

What is the set \(\{k_1,k_2,\ldots, k_i\}\cup \{j_1,j_2,\ldots, j_{n-i}\}\text{?}\)

In how many ways can you choose the set \(\{k_1,k_2,\ldots, k_i\}\text{?}\)

Once you have chosen this set, how many choices do you have for \(\{j_1,j_2,\ldots, j_{n-i}\}\text{?}\)

If you substitute \(x\) for each \(x_i\) and \(y\) for each \(y_i\text{,}\) how many terms of the form \(x^iy^{n-i}\) will you have in the expanded product

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n)=(x+y)^n? \end{equation*}

How many terms of the form \(x^{n-i}y^i\) will you have?

Explain how you have just proved your conjecture from Problem 53. The theorem you have proved is called the Binomial Theorem.

We can take this step-by-step and consider a small case to get started.

Now imagine expanding

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n). \end{equation*}

Once you apply the commutative law to the individual terms you get, you will have a sum of terms of the form

\begin{equation*} x_{k_1}x_{k_2}\cdots x_{k_i}\cdot y_{j_1}y_{j_2}\cdots y_{j_{n-i}}. \end{equation*}

What is the set \(\{k_1,k_2,\ldots, k_i\}\cup \{j_1,j_2,\ldots, j_{n-i}\}\text{?}\)

If you substitute \(x\) for each \(x_i\) and \(y\) for each \(y_i\text{,}\) how many terms of the form \(x^iy^{n-i}\) will you have in the expanded product

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n)=(x+y)^n? \end{equation*}

Now imagine expanding

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n). \end{equation*}

Once you apply the commutative law to the individual terms you get, you will have a sum of terms of the form

\begin{equation*} x_{k_1}x_{k_2}\cdots x_{k_i}\cdot y_{j_1}y_{j_2}\cdots y_{j_{n-i}}. \end{equation*}

What is the set \(\{k_1,k_2,\ldots, k_i\}\cup \{j_1,j_2,\ldots, j_{n-i}\}\text{?}\)

If you substitute \(x\) for each \(x_i\) and \(y\) for each \(y_i\text{,}\) how many terms of the form \(x^iy^{n-i}\) will you have in the expanded product

\begin{equation*} (x_1+y_1)(x_2+y_2)\cdots (x_n+y_n)=(x+y)^n? \end{equation*}

Explain how you have just proved your conjecture from Problem 53. The theorem you have proved is called the Binomial Theorem.

American coins are all marked with the year in which they were made. How many coins do you need to have in your hand to guarantee that on two (at least) of them, the date has the same last digit? (When we say "to guarantee that on two (at least) of them,..." we mean that you can find two with the same last digit. You might be able to find three with that last digit, or you might be able to find one pair with the last digit 1 and one pair with the last digit 9, or any combination of equal last digits, as long as there is at least one pair with the same last digit.)

Show that if \(S\) and \(T\) are finite sets of the same size, then a function \(f\) from \(S\) to \(T\) is one-to-one if and only if it is onto.

The previous exercise could help you prove that if \(f\) is one-to-one, then it is onto.

The sum principle can help you show that if \(f\) is an onto function, then \(f\) is one-to-one.

There is a generalized pigeonhole principle which says that if we partition a set with more than \(kn\) elements into \(n\) blocks, then at least one block has at least \(k+1\) elements. Prove the generalized pigeonhole principle.

Draw five circles labeled Al, Sue, Don, Pam, and Jo. Find a way to draw red and green lines between people so that every pair of people is joined by a line and there is neither a triangle consisting entirely of red lines or a triangle consisting of green lines. What does Problem 65 tell you about the possibility of doing this with six people's names? What does this problem say about the conclusion of Problem 65 holding when there are five people in our set rather than six?

Since \(R(3,3)=6\text{,}\) an uneducated guess might be that \(R(4,4)=8\text{.}\) Show that this is not the case.

What usually makes it hard for students to start this problem is the fact that we just defined what \(R(4, 4)\) is, and not what it means for a number not to be \(R(4, 4)\text{.}\) So to get started, try to write down what it means to say \(R(4, 4)\) is not 8. You will see that there are two things that can keep \(R(4, 4)\) from being 8. You need to figure out which one happens and explain why. One such explanation could involve the graph \(K_8\) .

Show that among ten people, there are either four mutual acquaintances or three mutual strangers. What does this say about \(R(4,3)\text{?}\)

Find \(R(4,3)\text{.}\)

In Problem 68 you showed that \(R(4, 3) \le 10\text{.}\) In Problem 70 you showed that \(R(4, 3) \gt 8\text{.}\) Thus \(R(4, 3)\) is either 9 or 10. Deciding which is the case is just plain hard. But there is a relevant problem we have done that we haven't used yet.

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CTGD Some Applications of the Basic Principles

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Problem 47 has a geometric interpretation in a coordinate plane. A lattice path in the plane is a "curve" made up of line segments that either go from a point \((i,j)\) to the point \((i+1,j)\) or from a point \((i,j)\) to the point \((i,j+1)\text{,}\) where \(i\) and \(j\) are integers.

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Let \(A\) be the dollar amount she gives to her older grandchild, and \(B\) be the dollar amount she gives to her younger grandchild. (As always, we start solving a word problem like this by defining the variables, including their units.)