1.1 Angles

Recall the following definitions from elementary geometry:

  1. An angle is acute if it is between \(0\Degrees\) and \(90\Degrees\).

  2. An angle is a right angle if it equals \(90\Degrees\).

  3. An angle is obtuse if it is between \(90\Degrees\) and \(180\Degrees\).

  4. An angle is a straight angle if it equals \(180\Degrees\).

tikzpicture [line width=0.5pt,-latex] (0.7,0) arc (0:45:0.7); [linecolor,line width=1.5pt,latex-latex] (1.5,0) -- (0,0) -- (1.3,1.3); [white] (1.6,0) -- (2.1,0); [white] (-0.1,0) -- (-0.5,0); tikzpicture

(a)  acute angle

tikzpicture [line width=0.5pt,-latex] (0.7,0) arc (0:90:0.7); [line width=0.5pt] (0.2,0) -- (0.2,0.2) -- (0,0.2); [linecolor,line width=1.5pt,latex-latex] (1.5,0) -- (0,0) -- (0,1.3); [white] (1.6,0) -- (2.1,0); [white] (-0.1,0) -- (-0.5,0); tikzpicture

(b)  right angle

tikzpicture [line width=0.5pt,-latex] (0.7,0) arc (0:135:0.7); [linecolor,line width=1.5pt,latex-latex] (1.5,0) -- (0,0) -- (-1.3,1.3); tikzpicture

(c)  obtuse angle

tikzpicture [line width=0.5pt,-latex] (0.7,0) arc (0:180:0.7); [linecolor,line width=1.5pt,latex-latex] (-1.5,0) -- (0,0) -- (1.5,0); tikzpicture

(d)  straight angle
Figure 1.1.1Types of angles

In elementary geometry, angles are always considered to be positive and not larger than \(360\Degrees\). For now we will only consider such angles.[1] The following definitions will be used throughout the text:

  1. Two acute angles are complementary if their sum equals \(90\Degrees\). In other words, if \(0\Degrees \le \angle\,A \,,\, \angle\,B \le 90\Degrees\) then \(\angle\,A\) and \(\angle\,B\) are complementary if \(\angle\,A + \angle\,B = 90\Degrees\).

  2. Two angles between \(0\Degrees\) and \(180\Degrees\) are supplementary if their sum equals \(180\Degrees\). In other words, if \(0\Degrees \le \angle\,A \,,\, \angle\,B \le 180\Degrees\) then \(\angle\,A\) and \(\angle\,B\) are supplementary if \(\angle\,A + \angle\,B = 180\Degrees\).

  3. Two angles between \(0\Degrees\) and \(360\Degrees\) are conjugate (or explementary) if their sum equals \(360\Degrees\). In other words, if \(0\Degrees \le \angle\,A \,,\, \angle\,B \le 360\Degrees\) then \(\angle\,A\) and \(\angle\,B\) are conjugate if \(\angle\,A + \angle\,B = 360\Degrees\).

tikzpicture [line width=0.5pt,latex-] (0,1) arc (90:56.3:1); [line width=0.5pt,-latex] (1,0) arc (0:56.3:1); [line width=0.5pt] (0.2,0) -- (0.2,0.2) -- (0,0.2); [linecolor,line width=1.5pt,latex-latex] (1.5,0) -- (0,0) -- (0,1.8); [linecolor,line width=1.5pt,-latex] (0,0) -- (1,1.5); [white] (1.6,0) -- (2.3,0); [white] (-0.1,0) -- (-0.7,0); [right] at (0.85,0.6) \,A; [above] at (0.35,1) \,B; tikzpicture

(a)  complementary

tikzpicture [line width=0.5pt,-latex] (0.7,0) arc (0:135:0.7); [line width=0.5pt,latex-] (-0.7,0) arc (180:135:0.7); [linecolor,line width=1.5pt,latex-latex] (-1.8,0) -- (1.8,0); [linecolor,line width=1.5pt,-latex] (0,0) -- (-1.3,1.3); [above right] at (0.1,0.8) \,A; [above left] at (-0.65,0.25) \,B; tikzpicture

(b)  supplementary

tikzpicture [line width=0.5pt,latex-] (0.7,0) arc (360:56.3:0.7); [line width=0.5pt,-latex] (0.7,0) arc (0:56.3:0.7); [linecolor,line width=1.5pt,latex-latex] (1.5,0) -- (0,0) -- (1,1.5); [right] at (0.8,0.5) \,A; [above] at (-0.7,0.5) \,B; tikzpicture

(c)  conjugate
Figure 1.1.2Types of pairs of angles

Instead of using the angle notation \(\angle\,A\) to denote an angle, we will sometimes use just a capital letter by itself (e.g. \(A\), \(B\), \(C\)) or a lowercase variable name (e.g. \(x\), \(y\), \(t\)). It is also common to use letters (either uppercase or lowercase) from the Greek alphabet, shown in the table below, to represent angles:

Table 1.1The Greek alphabet

LettersNameLettersNameLettersName
A\(\alpha\)alphaI\(\iota\)iotaP\(\rho\)rho
B\(\beta\)betaK\(\kappa\)kappa\(\Sigma\)\(\sigma\)sigma
\(\Gamma\)\(\gamma\)gamma\(\Lambda\)\(\lambda\)lambdaT\(\tau\)tau
\(\Delta\)\(\delta\)deltaM\(\mu\)mu\(\Upsilon\)\(\upsilon\)upsilon
E\(\epsilon\)epsilonN\(\nu\)nu\(\Phi\)\(\phi\)phi
Z\(\zeta\)zeta\(\Xi\)\(\xi\)xiX\(\chi\)chi
H\(\eta\)etaO\(o\)omicron\(\Psi\)\(\psi\)psi
\(\Theta\)\(\theta\)theta\(\Pi\)\(\pi\)pi\(\Omega\)\(\omega\)omega

In elementary geometry you learned that the sum of the angles in a triangle equals \(180\Degrees\), and that an isosceles triangle is a triangle with two sides of equal length. Recall that in a right triangle one of the angles is a right angle. Thus, in a right triangle one of the angles is \(90\Degrees\) and the other two angles are acute angles whose sum is \(90\Degrees\) (i.e. the other two angles are complementary angles).

Example 1.1

For each triangle below, determine the unknown angle(s):

tikzpicture [linecolor,line width=1.5pt,fill=fillcolor] (0,0) -- (4,0) -- (35:1.7) -- cycle; [below left] at (0,0) A; [above] at (35:1.7) B; [below right] at (4,0) C; at (15:0.75) 35 ; at (2.8,0.2) 20 ; [linecolor,line width=1.5pt,fill=fillcolor] (6,0) -- (8,0) -- (8,1.5) -- cycle; (7.8,0.025) -- (7.8,0.23) -- (7.975,0.23); [below left] at (6,0) D; [above right] at (8,1.5) E; [below right] at (8,0) F; at (7.7,0.9) 53 ; [linecolor,line width=1.5pt,fill=fillcolor] (10,0) -- (13,0) -- (11.5,1.3) -- cycle; [below left] at (10,0) X; [above] at (11.5,1.3) Y; [below right] at (13,0) Z; at (10.5,0.2) ; at (12.5,0.2) ; at (11.5,1) 3 ; tikzpicture

Note: We will sometimes refer to the angles of a triangle by their vertex points. For example, in the first triangle above we will simply refer to the angle \(\angle\,BAC\) as angle \(A\).

Solution: For triangle \(\triangle\,ABC\), \(A = 35\Degrees\) and \(C = 20\Degrees\), and we know that \(A + B + C = 180\Degrees\), so

\begin{displaymath} 35\Degrees ~+~ B ~+~ 20\Degrees ~=~ 180\Degrees \quad\Rightarrow\quad B ~=~ 180\Degrees ~-~ 35\Degrees ~-~ 20\Degrees \quad\Rightarrow\quad \boxed{B ~=~ 125\Degrees} ~. \end{displaymath}

For the right triangle \(\triangle\,DEF\), \(E = 53\Degrees\) and \(F = 90\Degrees\), and we know that the two acute angles \(D\) and \(E\) are complementary, so

\begin{displaymath} D ~+~ E ~=~ 90\Degrees \quad\Rightarrow\quad D ~=~ 90\Degrees ~-~ 53\Degrees \quad\Rightarrow\quad \boxed{D ~=~ 37\Degrees} ~. \end{displaymath}

For triangle \(\triangle\,XYZ\), the angles are in terms of an unknown number \(\alpha\), but we do know that \(X + Y + Z = 180\Degrees\), which we can use to solve for \(\alpha\) and then use that to solve for \(X\), \(Y\), and \(Z\):

\begin{displaymath} \alpha ~+~ 3\alpha ~+~ \alpha ~=~ 180\Degrees \quad\Rightarrow\quad 5\alpha ~=~ 180\Degrees \quad\Rightarrow\quad \alpha ~=~ 36\Degrees \quad\Rightarrow\quad \boxed{X = 36\Degrees ~,~ Y = 3 \times 36\Degrees = 108\Degrees ~,~ Z = 36\Degrees} \end{displaymath}

Example 1.2

Thales’ Theorem states that if \(A\), \(B\), and \(C\) are (distinct) points on a circle such that the line segment \(\overline{AB}\) is a diameter of the circle, then the angle \(\angle\,ACB\) is a right angle (see Figure 1.1.3(a)). In other words, the triangle \(\triangle\,ABC\) is a right triangle.

tikzpicture[scale=0.8] [linecolor,fill=fillcolor,line width=1.5pt] (-2,0) -- (2,0) -- (70:2) -- (-2,0); [line width=0.5pt] (0,0) circle (2); [left] at (-2,0) A; [right] at (2,0) B; [above] at (70:2) C; [below] at (0,0) O; (-2,0) circle (2pt); (2,0) circle (2pt); (70:2) circle (2pt); (0,0) circle (2pt); tikzpicture

tikzpicture[scale=0.8] [linecolor,fill=fillcolor,line width=1.5pt] (-2,0) -- (2,0) -- (70:2) -- (-2,0); [line width=0.5pt] (0,0) circle (2); [line width=0.5pt] (0,0) -- (70:2); [left] at (-2,0) A; [right] at (2,0) B; [above] at (70:2) C; [below] at (0,0) O; at (-1.3,0.2) ; at (77:1.4) ; at (62:1.52) ; at (1.5,0.2) ; (-2,0) circle (2pt); (2,0) circle (2pt); (70:2) circle (2pt); (0,0) circle (2pt); tikzpicture

Figure 1.1.3Thales’ Theorem: \(\angle\,ACB = 90\Degrees\)

To prove this, let \(O\) be the center of the circle and draw the line segment \(\overline{OC}\), as in Figure 1.1.3(b). Let \(\alpha = \angle\,BAC\) and \(\beta = \angle\,ABC\). Since \(\overline{AB}\) is a diameter of the circle, \(\overline{OA}\) and \(\overline{OC}\) have the same length (namely, the circle’s radius). This means that \(\triangle\,OAC\) is an isosceles triangle, and so \(\angle\,OCA = \angle\,OAC = \alpha\). Likewise, \(\triangle\,OBC\) is an isosceles triangle and \(\angle\,OCB = \angle\,OBC = \beta\). So we see that \(\angle\,ACB = \alpha + \beta\). And since the angles of \(\triangle\,ABC\) must add up to \(180\Degrees\), we see that \(180\Degrees = \alpha + ( \alpha + \beta ) + \beta = 2\,( \alpha + \beta )\), so \(\alpha + \beta = 90\Degrees\). Thus, \(\angle\,ACB = 90\Degrees\). \(\text{qed}\)

tikzpicture[scale=0.5] [fill=fillcolor] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.6,0) -- (3.6,0.4) -- (4,0.4); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) A; [below right] at (4,0) C; [above right] at (4,3) B; [below] at (2,0) b; [right] at (4,1.5) a; [above left] at (2,1.5) c; tikzpicture

Figure 1.1.4

In a right triangle, the side opposite the right angle is called the hypotenuse, and the other two sides are called its legs. For example, in Figure 1.1.4 the right angle is \(C\), the hypotenuse is the line segment \(\overline{AB}\), which has length \(c\), and \(\overline{BC}\) and \(\overline{AC}\) are the legs, with lengths \(a\) and \(b\), respectively. The hypotenuse is always the longest side of a right triangle (see Exercise 11).

By knowing the lengths of two sides of a right triangle, the length of the third side can be determined by using the Pythagorean Theorem:

Theorem 1.1

Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of its legs.

Thus, if a right triangle has a hypotenuse of length \(c\) and legs of lengths \(a\) and \(b\), as in Figure 1.1.4, then the Pythagorean Theorem says:

\begin{equation} {\setlength\fboxsep{2mm}\setlength\fboxrule{1pt}\boxed{a^2 ~+~ b^2 ~=~ c^2}} \tag{1.1}\end{equation}

Let us prove this. In the right triangle \(\triangle\,ABC\) in Figure 1.1.5(a) below, if we draw a line segment from the vertex \(C\) to the point \(D\) on the hypotenuse such that \(\overline{CD}\) is perpendicular to \(\overline{AB}\) (that is, \(\overline{CD}\) forms a right angle with \(\overline{AB}\)), then this divides \(\triangle\,ABC\) into two smaller triangles \(\triangle\,CBD\) and \(\triangle\,ACD\), which are both similar to \(\triangle\,ABC\).

tikzpicture[scale=0.67] [fill=yellow!30] (0,0) -- (4,0) -- (2.56,1.92) -- (0,0); [fill=green!40] (4,0) -- (2.56,1.92) -- (4,3) -- (4,0); [line width=0.5pt] (3.7,0) -- (3.7,0.3) -- (4,0.3); [line width=0.5pt,rotate=36.87] (2.9,0) -- (2.9,-0.3) -- (3.5,-0.3) -- (3.5,0); [dashed,line width=0.5pt] (4,0) -- (2.56,1.92); [black!60,line width=0.5pt] (1,0) arc (0:36.87:1); [black!60,line width=0.5pt,rotate=18.435] (0.8,0) -- (1.2,0); [black!60,line width=0.5pt] (4,2.2) arc (270:216.87:0.8); [black!60,line width=0.5pt,rotate around=-22.565:(4,3)] (4,2.4) -- (4,2); [black!60,line width=0.5pt,rotate around=-30.565:(4,3)] (4,2.4) -- (4,2); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) A; [below right] at (4,0) C; [above right] at (4,3) B; [below] at (2,0) b; [right] at (4,1.5) a; [above left] at (1.35,2.45) c; [snake=brace,segment amplitude=3mm,rotate=36.87] (0,0.65) -- (5,0.65); [above] at (2.45,1.87) D; [above,rotate=36.87] at (3.28,2.46) d; [above,rotate=36.87] at (1.28,0.96) c-d; tikzpicture

(a)  \(\triangle\,ABC\)

tikzpicture[scale=0.402] [fill=green!40] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.5,0) -- (3.5,0.5) -- (4,0.5); [black!60,line width=0.5pt] (4,1.8) arc (270:216.87:1.2); [black!60,line width=0.5pt,rotate around=-22.565:(4,3)] (4,2.2) -- (4,1.4); [black!60,line width=0.5pt,rotate around=-30.565:(4,3)] (4,2.2) -- (4,1.4); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) C; [below right] at (4,0) D; [above right] at (4,3) B; [right] at (4,1.5) d; [above left] at (2,1.5) a; tikzpicture

(b)  \(\triangle\,CBD\)

tikzpicture[scale=0.536] [fill=yellow!30] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.625,0) -- (3.625,0.375) -- (4,0.375); [black!60,line width=0.5pt] (1.3,0) arc (0:36.87:1.3); [black!60,line width=0.5pt,rotate=18.435] (1.1,0) -- (1.5,0); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) A; [below right] at (4,0) D; [above right] at (4,3) C; [below] at (2,0) c-d; [above left] at (2,1.5) b; tikzpicture

(c)  \(\triangle\,ACD\)
Figure 1.1.5Similar triangles \(\triangle\,ABC\), \(\triangle\,CBD\), \(\triangle\,ACD\)

Recall that triangles are similar if their corresponding angles are equal, and that similarity implies that corresponding sides are proportional. Thus, since \(\triangle\,ABC\) is similar to \(\triangle\,CBD\), by proportionality of corresponding sides we see that

\begin{displaymath} \overline{AB}~\text{is to}~\overline{CB}~\text{(hypotenuses)}\enskip\text{as}\enskip \overline{BC}~\text{is to}~\overline{BD}~\text{(vertical legs)} \quad\Rightarrow\quad \frac{c}{a} ~=~ \frac{a}{d} \quad\Rightarrow\quad cd ~=~ a^2 ~. \end{displaymath}

Since \(\triangle\,ABC\) is similar to \(\triangle\,ACD\), comparing horizontal legs and hypotenuses gives

\begin{displaymath} \frac{b}{c-d} ~=~ \frac{c}{b} \quad\Rightarrow\quad b^2 ~=~ c^2 ~-~ cd ~=~ c ^2 ~-~ a^2 \quad\Rightarrow\quad a^2 ~+~ b^2 ~=~ c^2 ~. \text{qed} \end{displaymath}

Note: The symbols \(\perp\) and \(\sim\) denote perpendicularity and similarity, respectively. For example, in the above proof we had \(\,\overline{CD} \perp \overline{AB}\,\) and \(\,\triangle\,ABC \sim \triangle\,CBD \sim \triangle\,ACD\).

Example 1.3

For each right triangle below, determine the length of the unknown side:

tikzpicture[scale=0.4,every node/.style=font=] [fill=fillcolor] (0,0) -- (4,0) -- (4,3) -- (0,0); [line width=0.5pt] (3.625,0) -- (3.625,0.375) -- (4,0.375); [linecolor,line width=1.5pt] (0,0) -- (4,0) -- (4,3) -- cycle; [below left] at (0,0) A; [below right] at (4,0) C; [above right] at (4,3) B; [below] at (2,0) 4; [right] at (4,1.5) a; [above left] at (2,1.5) 5; [fill=fillcolor] (8,0) -- (10.732,0) -- (10.732,2) -- (8,0); [line width=0.5pt] (10.357,0) -- (10.357,0.375) -- (10.732,0.375); [linecolor,line width=1.5pt] (8,0) -- (10.732,0) -- (10.732,2) -- cycle; [below left] at (8,0) D; [below right] at (10.732,0) F; [above right] at (10.732,2) E; [below] at (9.366,0) e; [right] at (10.732,1) 1; [above left] at (9.366,1) 2; [fill=fillcolor] (15,0) -- (18,0) -- (18,3) -- (15,0); [line width=0.5pt] (17.625,0) -- (17.625,0.375) -- (18,0.375); [linecolor,line width=1.5pt] (15,0) -- (18,0) -- (18,3) -- cycle; [below left] at (15,0) X; [below right] at (18,0) Z; [above right] at (18,3) Y; [below] at (16.5,0) 1; [right] at (18,1.5) 1; [above left] at (16.5,1.5) z; tikzpicture

Solution: For triangle \(\triangle\,ABC\), the Pythagorean Theorem says that

\begin{displaymath} a^2 ~+~ 4^2 ~=~ 5^2 \quad\Rightarrow\quad a^2 ~=~ 25 ~-~ 16 ~=~ 9 \quad\Rightarrow\quad \boxed{a ~=~ 3} ~. \end{displaymath}

For triangle \(\triangle\,DEF\), the Pythagorean Theorem says that

\begin{displaymath} e^2 ~+~ 1^2 ~=~ 2^2 \quad\Rightarrow\quad e^2 ~=~ 4 ~-~ 1 ~=~ 3 \quad\Rightarrow\quad \boxed{e ~=~ \sqrt{3}} ~. \end{displaymath}

For triangle \(\triangle\,XYZ\), the Pythagorean Theorem says that

\begin{displaymath} 1^2 ~+~ 1^2 ~=~ z^2 \quad\Rightarrow\quad z^2 ~=~ 2 \quad\Rightarrow\quad \boxed{z ~=~ \sqrt{2}} ~. \end{displaymath}

Example 1.4

[r]tikzpicture[scale=0.8] [brickcolor] (1.25,0) -- (1.25,3.3) -- (2,3.3) -- (2,0) -- (1.25,0); [pattern color=white,pattern=bricks] (1.25,0) -- (1.25,3.3) -- (2,3.3) -- (2,0) -- (1.25,0); [black!10] (-0.7,0) -- (2,0) -- (2,-0.8) -- (-0.7,-0.8) -- (-0.7,0); [black!60] (-0.2,0) -- (1.25,3); [line width=2pt] (-0.7,0) -- (2,0); [line width=2pt] (1.25,0) -- (1.25,3.3); scope[>=latex] [<->|] (1.57,0) -- (1.57,2.93) node [midway,right] h; [|<->|] (-0.2,-0.3) -- (1.25,-0.3) node [midway,below] 8; scope [left] at (0.615,1.5) 17; at (0.85,0.25) 90 ; tikzpicture A 17 ft ladder leaning against a wall has its foot 8 ft from the base of the wall. At what height is the top of the ladder touching the wall?

Solution: Let \(h\) be the height at which the ladder touches the wall. We can assume that the ground makes a right angle with the wall, as in the picture on the right. Then we see that the ladder, ground, and wall form a right triangle with a hypotenuse of length 17 ft (the length of the ladder) and legs with lengths 8 ft and \(h\) ft. So by the Pythagorean Theorem, we have

\begin{displaymath} h^2 ~+~ 8^2 ~=~ 17^2 \quad\Rightarrow\quad h^2 ~=~ 289 ~-~ 64 ~=~ 225 \quad\Rightarrow\quad \boxed{h ~=~ 15 ~\text{ft}} ~. \end{displaymath}

Exercises

For Exercises 1-4, find the numeric value of the indicated angle(s) for the triangle \(\triangle\,ABC\).

2

  1. Find \(B\) if \(A = 15\Degrees\) and \(C = 50\Degrees\).

  2. Find \(C\) if \(A = 110\Degrees\) and \(B = 31\Degrees\).

    2

  3. Find \(A\) and \(B\) if \(C = 24\Degrees\), \(A = \alpha\), and \(B = 2\alpha\).

  4. Find \(A\), \(B\), and \(C\) if \(A = \beta\) and \(B = C = 4\beta\).

For Exercises 5-8, find the numeric value of the indicated angle(s) for the right triangle \(\triangle\,ABC\), with \(C\) being the right angle.

2

  1. Find \(B\) if \(A = 45\Degrees\).

  2. Find \(A\) and \(B\) if \(A = \alpha\) and \(B = 2\alpha\).

    2

  3. Find \(A\) and \(B\) if \(A = \phi\) and \(B = \phi^2\).

  4. Find \(A\) and \(B\) if \(A = \theta\) and \(B = 1/\theta\).

  5. A car goes 24 miles due north then 7 miles due east. What is the straight distance between the car’s starting point and end point?

  6. One end of a rope is attached to the top of a pole 100 ft high. If the rope is 150 ft long, what is the maximum distance along the ground from the base of the pole to where the other end can be attached? You may assume that the pole is perpendicular to the ground.

  7. Prove that the hypotenuse is the longest side in every right triangle. (Hint: Is \(a^2 + b^2 > a^2\)?)

  8. Can a right triangle have sides with lengths 2, 5, and 6? Explain your answer.

  9. If the lengths \(a\), \(b\), and \(c\) of the sides of a right triangle are positive integers, with \(a^2 + b^2 = c^2\), then they form what is called a Pythagorean triple. The triple is normally written as (\(a\),\(b\),\(c\)). For example, (3,4,5) and (5,12,13) are well-known Pythagorean triples.

    1. Show that (6,8,10) is a Pythagorean triple.

    2. Show that if (\(a\),\(b\),\(c\)) is a Pythagorean triple then so is (\(ka\),\(kb\),\(kc\)) for any integer \(k >0\). How would you interpret this geometrically?

    3. Show that (\(2mn\),\(m^2 - n^2\),\(m^2 + n^2\)) is a Pythagorean triple for all integers \(m > n > 0\).

    4. The triple in part(c) is known as Euclid’s formula for generating Pythagorean triples. Write down the first ten Pythagorean triples generated by this formula, i.e. use: \(m=2\) and \(n=1\); \(m=3\) and \(n=1\), \(2\); \(m=4\) and \(n=1\), \(2\), \(3\); \(m=5\) and \(n=1\), \(2\), \(3\), \(4\).

  10. This exercise will describe how to draw a line through any point outside a circle such that the line intersects the circle at only one point. This is called a tangent line to the circle (see the picture on the left in Figure 1.1.6), a notion which we will use throughout the text.

    tikzpicture (0,0) circle (1.5); [line width=0.7pt,latex-] (-2.5,1.5) -- (3.5,1.5) node [pos=0.7,above] tangent line; (0,1.5) circle (2pt); (3.5,1.5) circle (2pt); [line width=0.7pt,-latex] (3.5,1.5) -- (-2.5,-0.7) node [pos=0.6,below,sloped] not tangent; tikzpicture tikzpicture[scale=0.65] (a) at (2in,0in); [circle,draw] (c) at (0in,0in) [minimum size=1in] ; [linecolor,fill=fillcolor,fill opacity=0.4,line width=1.5pt] (a) -- (tangent cs:node=c,point=(a),solution=1) node [black,fill opacity=1] node [above,black,fill opacity=1] A -- (c.center) -- (a); (1in,0in) circle (1in); (0in,0in) circle (3pt); (1in,0in) circle (3pt); (2in,0in) circle (3pt); [left] at (0in,0in) O; [right] at (2in,0in) P; [below] at (1in,0in) C; tikzpicture

    inside

    Figure 1.1.6


    On a sheet of paper draw a circle of radius 1 inch, and call the center of that circle \(O\). Pick a point \(P\) which is \(2.5\) inches away from \(O\). Draw the circle which has \(\overline{OP}\) as a diameter, as in the picture on the right in Figure 1.1.6. Let \(A\) be one of the points where this circle intersects the first circle. Draw the line through \(P\) and \(A\). In general the tangent line through a point on a circle is perpendicular to the line joining that point to the center of the circle (why?). Use this fact to explain why the line you drew is the tangent line through \(A\) and to calculate the length of \(\overline{PA}\). Does it match the physical measurement of \(\overline{PA}\)? [r]tikzpicture[scale=0.68] [linecolor,fill=fillcolor,line width=1.5pt] (-2,0) -- (2,0) -- (70:2) -- (-2,0); [line width=0.5pt] (0,0) circle (2); [left] at (-2,0) A; [right] at (2,0) B; [above] at (70:2) C; [below] at (0,0) O; [dashed] (-2,0) -- (250:2); [dashed] (250:2) -- (2,0); (-2,0) circle (2.9pt); (2,0) circle (2.9pt); (70:2) circle (2.9pt); (0,0) circle (2.9pt); tikzpicture

  11. Suppose that \(\triangle\,ABC\) is a triangle with side \(\overline{AB}\) a diameter of a circle with center \(O\), as in the picture on the right, and suppose that the vertex \(C\) lies on the circle. Now imagine that you rotate the circle \(180\Degrees\) around its center, so that \(\triangle\,ABC\) is in a new position, as indicated by the dashed lines in the picture. Explain how this picture proves Thales’ Theorem.

  1. Later in the text we will discuss negative angles and angles larger than \(360\Degrees\). ↩