Chapter 1: Right Triangles and an Introduction to Trigonometry
Chapter Outline
 1.1. The Pythagorean Theorem
 1.2. Special Right Triangles
 1.3. Basic Trigonometric Functions
 1.4. Solving Right Triangles
 1.5. Measuring Rotation
 1.6. Applying Trig Functions to Angles of Rotation
 1.7. Trigonometric Functions of Any Angle
 1.8. Relating Trigonometric Functions
Chapter Summary
Chapter Summary
In this chapter students learned about right triangles and special right triangles. Through the special right triangles and the Pythagorean Theorem, the study of trigonometry was discovered. Sine, cosine, tangent, secant, cosecant, and cotangent are all functions of angles and the result is the ratio of the sides of a right triangle. We learned that only our special right triangles generate sine, cosine, tangent values that can be found without the use of a scientific calculator. When incorporating the trig ratios and the Pythagorean Theorem, we discovered the first of many trig identities. This concept will be explored further in Chapter 3.
Vocabulary
 Adjacent
 A side adjacent to an angle is the side next to the angle. In a right triangle, it is the leg that is next to the angle.
 Angle of depression
 The angle between the horizontal line of sight, and the line of sight down to a given point.
 Angle of elevation
 The angle between the horizontal line of sight, and the line of sight up to a given point.
 Bearings
 The direction from one object to another, usually measured as an angle.
 Clinometer
 A device used to measure angles of elevation or depression.
 Coterminal angles
 Two angles in standard position are coterminal if they share the same terminal side.
 Distance Formula

\begin{align*}d = \sqrt{(x_1x_2)^2 + (y_1y_2)^2}\end{align*}
d=(x1−x2)2+(y1−y2)2−−−−−−−−−−−−−−−−−−√
 Hypotenuse
 The hypotenuse is the longest side in a right triangle, opposite the right angle.
 Identity
 An identity is an equation that is always true, as long as the variables and expressions involved are defined.
 Included Angle
 The angle inbetween two sides of a polygon.
 Leg
 The legs of a right triangle are the two shorter sides.
 Nautical Mile
 A nautical mile is a unit of length that corresponds approximately to one minute of latitude along any meridian. A nautical mile is equal to 1.852 meters.
 Pythagorean Theorem

\begin{align*}a^2 + b^2 = c^2\end{align*}
a2+b2=c2
 Pythagorean Triple
 A set whole numbers for which the Pythagorean Theorem holds true.
 Quadrantal angle
 A quadrantal angle is an angle in standard position whose terminal side lies on an axis.
 Radius
 The radius of a circle is the distance from the center of the circle to the edge. The radius defines the circle.
 Reference angle

The reference angle of an angle in standard position is the measure of the angle between the terminal side and the closest portion of the \begin{align*}x\end{align*}
x− axis.
 Standard position

An angle in standard position has its initial side on the positive \begin{align*}x\end{align*}
x− axis, its vertex at the origin, and its terminal side anywhere in the plane. A positive angle means a counterclockwise rotation. A negative angle means a clockwise rotation.
 Theodolite
 A device used to measure angles of elevation or depression.
 Unit Circle

The unit circle is the circle with radius 1 and center (0, 0). The equation of the unit circle is \begin{align*}x^2 + y^2 = 1\end{align*}
x2+y2=1
Review Questions
 One way to prove the Pythagorean Theorem is through the picture below. Determine the area of the square two different ways and set each equal to each other.
 A flute is resting diagonally, \begin{align*}d\end{align*}
d , in the rectangular box (prism) below. Find the length of the flute.  Solve the right triangle.
 Solve the right triangle.
 Find the exact value of the area of the parallelogram below.
 The modern building shown below is built with an outer wall (shown on the left) that is not at a 90degree angle with the floor. The wall on the right is perpendicular to both the floor and ceiling. Find the length of \begin{align*}w\end{align*}
w .  Given that \begin{align*}\cos(90^\circx) = \frac{2}{7}\end{align*}
cos(90∘−x)=27 , find the \begin{align*}\sin x\end{align*}sinx .  If \begin{align*}\cos(x) = \frac{3}{4}\end{align*}
cos(−x)=34 and \begin{align*}\tan x = \frac{\sqrt{7}}{3}\end{align*}tanx=7√3 , find \begin{align*}\sin(x)\end{align*}sin(−x) .  If \begin{align*}\sin y = \frac{1}{3}\end{align*}
siny=13 , what is \begin{align*}\cos y\end{align*}cosy ? 
\begin{align*}\sin \theta = \frac{1}{3}\end{align*}
sinθ=13 find the value(s) of \begin{align*}\cos \theta\end{align*}cosθ . 
\begin{align*}\cos \theta = \frac{2}{5}\end{align*}
cosθ=−25 , and \begin{align*}\theta\end{align*}θ is a second quadrant angle. Find the exact values of remaining trigonometric functions.  (3, 4) is a point on the terminal side of \begin{align*}\theta\end{align*}
θ . Find the exact values of the six trigonometric functions.  Determine reference angle and two coterminal angles for \begin{align*}165^\circ\end{align*}
165∘ . Plot the angle in standard position.  It is very helpful to have the unit circle with all the special values on one circle. Fill out the unit circle below with all of the endpoints for each special value and quadrantal value.
Review Answers
 Area 1: \begin{align*}& (a+b)^2\\
& (a+b)(a+b)\\
& a^2+2ab+b^2\end{align*}Area 2: Add up 4 triangles and inner square. \begin{align*}4 \cdot \frac{1}{2}ab + c^2\\ 2ab + c^2\end{align*}
(a+b)2(a+b)(a+b)a2+2ab+b2 Set the two equal to each other: \begin{align*}a^2+2ab+b^2 & =2ab+c^2\\ a^2+b^2 & = c^2\end{align*}4⋅12ab+c22ab+c2 a2+2ab+b2a2+b2=2ab+c2=c2  First, find the diagonal of the base. This is a Pythagorean Triple, so the base diagonal is 25 (you could have also done the Pythagorean Theorem if you didn’t see this). Now, do the Pythagorean Theorem with the height and the diagonal to get the threedimensional diagonal. \begin{align*}7^2 + 25^2 & = d^2\\
49 + 225 & = d^2\\
274 & = d^2\\
\sqrt{274} & = d \approx 16.55\end{align*}
72+25249+225274274−−−√=d2=d2=d2=d≈16.55 
\begin{align*}\angle{C} = 90^\circ23.6^\circ = 66.4^\circ\\
\sin 23.6 & = \frac{CA}{25} \qquad \qquad \quad \ \cos 23.6 = \frac{AT}{25}\\
25 \cdot \sin 23.6 & = CA \qquad \qquad 25 \cdot \cos 23.6 = AT\\
10.01 & \approx CA \qquad \qquad \qquad \quad \ 22.9 \approx AT\end{align*}
∠C=90∘−23.6∘=66.4∘sin23.625⋅sin23.610.01=CA25 cos23.6=AT25=CA25⋅cos23.6=AT≈CA 22.9≈AT  First do the Pythagorean Theorem to get the third side. \begin{align*}7^2 + x^2 & = 18^2\\
49 + x^2 & =324\\
x^2 & = 275\\
x & = \sqrt{275} = 5\sqrt{11}\end{align*}Second, use one of the inverse functions to find the two missing angles. \begin{align*}\sin G & = \frac{7}{18}\\ \sin^{1} \left ( \frac{7}{18} \right ) & = G && \text{We can subtract} \ \angle{G} \ \text{from 90 to get} \ 67.11^\circ.\\ G & \approx 22.89^\circ\end{align*}
72+x249+x2x2x=182=324=275=275−−−√=511−−√ sinGsin−1(718)G=718=G≈22.89∘We can subtract ∠G from 90 to get 67.11∘. 
\begin{align*}A & = ab \sin C\\
& = 16 \cdot 22 \cdot \sin 60^\circ\\
& = 352 \cdot \frac{\sqrt{3}}{2}\\
& = 176 \sqrt{3}\end{align*}
 Make a right triangle with 165 as the opposite leg and \begin{align*}w\end{align*} is the hypotenuse. \begin{align*}\sin 85^\circ & = \frac{165}{w}\\
w\sin 85^\circ & = 165\\
w & = \frac{165}{\sin 85^\circ}\\
w & \approx 165.63\end{align*}

\begin{align*}\cos(90^\circx) & = \sin x\\
\sin x & = \frac{2}{7}\end{align*}
 If \begin{align*}\cos(x)= \frac{3}{4}\end{align*}, then \begin{align*}\cos x = \frac{3}{4}\end{align*}. With \begin{align*}\tan x = \frac{\sqrt{7}}{3}\end{align*}, we can conclude that \begin{align*}\sin x = \frac{\sqrt{7}}{4}\end{align*} and \begin{align*}\sin(x) = \frac{\sqrt{7}}{4}\end{align*}.
 If \begin{align*}\sin y = \frac{1}{3}\end{align*}, then we know the opposite side and the hypotenuse. Using the Pythagorean Theorem, we get that the adjacent side is \begin{align*}2\sqrt{2} \left (1^2 + b^2 = 3^2 \rightarrow b = \sqrt{91} \rightarrow b = \sqrt{8} = 2 \sqrt{2} \right )\end{align*}. Thus, \begin{align*}\cos y = \pm \frac{2\sqrt{2}}{3}\end{align*} because we don’t know if the angle is in the second or third quadrant.

\begin{align*}\sin \theta = \frac{1}{3}\end{align*}, sine is positive in Quadrants I and II. So, there can be two possible answers for the \begin{align*}\cos \theta\end{align*}. Find the third side, using the Pythagorean Theorem: \begin{align*}1^2 + b^2 & = 3^2\\
1 + b^2 & = 9\\
b^2 & = 8\\
b & = \sqrt{8} = 2\sqrt{2}\end{align*}In Quadrant I, \begin{align*}\cos \theta = \frac{2\sqrt{2}}{3}\end{align*} In Quadrant II, \begin{align*}\cos \theta = \frac{2\sqrt{2}}{3}\end{align*}

\begin{align*}\cos \theta = \frac{2}{5}\end{align*} and is in Quadrant II, so from the Pythagorean Theorem : \begin{align*}a^2 + (2)^2 & = 5^2\\
a^2 + 4 & = 25\\
a^2 & = 21\\
a & = \sqrt{21}\end{align*}So, \begin{align*}\sin \theta = \frac{\sqrt{21}}{5}\end{align*} and \begin{align*}\tan \theta = \frac{\sqrt{21}}{2}\end{align*}
 If the terminal side of \begin{align*}\theta\end{align*} is on (3, 4) means \begin{align*}\theta\end{align*} is in Quadrant IV, so cosine is the only positive function. Because the two legs are lengths 3 and 4, we know that the hypotenuse is 5. 3, 4, 5 is a Pythagorean Triple (you can do the Pythagorean Theorem to verify). Therefore, \begin{align*}\sin \theta = \frac{3}{5}, \cos \theta = \frac{4}{5}, \tan \theta = \frac{4}{3}\end{align*}
 Reference angle \begin{align*}= 15^\circ\end{align*}. Possible coterminal angles \begin{align*}= 195^\circ, 525^\circ\end{align*}
Texas Instruments Resources
In the CK12 Texas Instruments Trigonometry FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9699.