Trigonometry, Part I Tom Lewis Spring Semester 2011 Tom Lewis () Trigonometry, Part I Spring Semester 2011 1 / 24 Outline 1 Angle and directed angle 2 Angle measure 3 The length of an arc of a circle 4 Some geometry 5 Right-triangle trigonometry 6 Solving non-right triangles Tom Lewis () Trigonometry, Part I Spring Semester 2011 2 / 24 Angle and directed angle Angle and directed angle Here are three representations of \ABC. The first we might call static. In the second example, we imagine the angle being swept out in the counterclockwise direction; we will call this a positive directed angle. In the third example, we imagine the angle being swept out in the clockwise direction; we will call this a negative directed angle. In each case we call B the vertex of the angle. Tom Lewis () Trigonometry, Part I Spring Semester 2011 3 / 24 Angle and directed angle Standard Position We will often embed directed angles within a Cartesian coordinate system. An angle is in standard position when the vertex of the angle is at the origin and its initial side coincides with the positive x-axis. The figure above shows a positive and and negative directed angles in standard position. Tom Lewis () Trigonometry, Part I Spring Semester 2011 4 / 24 Angle measure Degree measure In degree measure, the full circle is divided into 360 equal angles; each division gives one degree. Angles can be measured by a protractor. Tom Lewis () Trigonometry, Part I Spring Semester 2011 5 / 24 Angle measure Problem Graph the following angles in standard position. 60◦ −135◦ 440◦ −580◦ Tom Lewis () Trigonometry, Part I Spring Semester 2011 6 / 24 Angle measure Radian measure When measuring in radians, the vertex of the angle is placed at the center of a unit circle (a circle of radius 1). The radian measure of the angle is the length of the arc subtended by the angle. Tom Lewis () Trigonometry, Part I Spring Semester 2011 7 / 24 Angle measure The use of symmetry The unit circle has a circumference of 2π. The radian measure of various angles can thus be found by symmetry. For example, we can divide the unit circle into 8 equal parts; thus, each part will have a radian measure of 2π=8 = π=4. Tom Lewis () Trigonometry, Part I Spring Semester 2011 8 / 24 Angle measure Some additional elementary angles If we divide the unit circle into 12 equal parts, then each part has a radian measure of 2π=12 = π=6. We can lay out these angles and their radian measures in standard position. Tom Lewis () Trigonometry, Part I Spring Semester 2011 9 / 24 Angle measure Conversion We can make conversions between degree to radian measure by observing that 180◦ = π rad For example, π 8 96◦ = 96 = rad 180 15 Likewise 180 5 rad = 5 ≈ 286:479◦ π Tom Lewis () Trigonometry, Part I Spring Semester 2011 10 / 24 Angle measure Problem Convert the following from degrees to radians or from radians to degrees: 72◦ −100◦ 20π=3 rad 7:5 rad Tom Lewis () Trigonometry, Part I Spring Semester 2011 11 / 24 Angle measure Solution ◦ 2 72 = 5 π rad ≈ 1:257 rad ◦ 5 −100 = − 9 π rad ≈ −1:745 rad 20π=3 rad = 1200◦ 7:5 rad = 429:718◦ Tom Lewis () Trigonometry, Part I Spring Semester 2011 12 / 24 The length of an arc of a circle Theorem If the arc of a circle of radius r subtends an angle of radian measure a, then the length s of the arc is given by s = ra: Tom Lewis () Trigonometry, Part I Spring Semester 2011 13 / 24 The length of an arc of a circle Problem Find the length of the arc of a circle of radius 15ft. subtending an angle of 72◦. Solution First we must convert 72◦ to radian measure: 72◦ = 72π=180 ≈ 1:257 rad. To obtain the length of the arc, we use the formula s = ra. In this case s ≈ (15)(1:257) = 18:855ft. Tom Lewis () Trigonometry, Part I Spring Semester 2011 14 / 24 Some geometry Theorem The interior angles of a triangle sum to 180◦ or π rad. Problem Find the angle between consecutive sides of a regular pentagon. Solution The answer is 108◦. Tom Lewis () Trigonometry, Part I Spring Semester 2011 15 / 24 Right-triangle trigonometry Conventions In a triangle, we will assume the following conventions: the angles will be denoted by α, β, and γ, and the the sides opposite these angles by a, b, and c, respectively. In a right triangle, we will assume that γ is the right angle. The side opposite the right angle is called the hypotenuse. Tom Lewis () Trigonometry, Part I Spring Semester 2011 16 / 24 Right-triangle trigonometry Definition To solve a triangle means to determine all of its angles and sides. Solving triangle requires the knowledge of the trigonometric functions: sin(α) = a=c cos(α) = b=c tan(α) = a=b csc(α) = c=a sec(α) = c=b cot(α) = b=a Problem Show that all 6 trigonometric functions can be expressed in terms of sine and cosine alone. Tom Lewis () Trigonometry, Part I Spring Semester 2011 17 / 24 Right-triangle trigonometry Problem Give the exact solution of the right triangle with c = 1, α = 30◦ and β = 60◦. Give the exact solution of the right triangle with c = 1, α = 45◦ and β = 45◦. Solution Using properties of the equilateral triangle andp the Pythagorean theorem, we can show that a = 1=2 and b = 3=2. Using properties of isoscelesp triangles and the Pythagorean theorem, we can show that a = b = 2=2. Tom Lewis () Trigonometry, Part I Spring Semester 2011 18 / 24 Right-triangle trigonometry Problem Solve the right triangle with c = 10 and α = 32 deg. Solve the right triangle with a = 8 and β = 40 deg. Solve the triangle with a = 5 and b = 12. Solution a = 10 sin(32) ≈ 5:299; b = 10 sin(58) ≈ 8:48. α = 50◦; c = 8= sin(50) ≈ 10:44; b = c sin(40) ≈ 6:71. p c = 52 + 122 = 13; α = sin−1(5=13) ≈ 22:62◦; β ≈ 67:38◦ Tom Lewis () Trigonometry, Part I Spring Semester 2011 19 / 24 Right-triangle trigonometry Problem A 6 foot man is 50 feet from the base of a tree. If the angle of elevation of the top of the tree is 40◦, then how tall is the tree? Solution If h is the height of the tree, then it is not hard to see that h = 6 + 50 tan(40) ≈ 47:95 ft. Tom Lewis () Trigonometry, Part I Spring Semester 2011 20 / 24 Solving non-right triangles Two laws There are two important tools used in solving non-right triangles: the law of sines and the law of cosines. The law of sines says that in any triangle sin(α) sin(β) sin(γ) = = a b c The law of cosines states that c2 = a2 + b2 − 2ab cos(γ) This identity can be expressed with the roles of a, b, and c (hence α, β, and γ) exchanged. Tom Lewis () Trigonometry, Part I Spring Semester 2011 21 / 24 Solving non-right triangles A strategy Here is a general rule: if you are given two angles and a side, then first use the law of sines; if you are given at least two sides (and perhaps an angle), then first use the law of cosines. Problem Solve the triangle with β = 33◦, γ = 58◦, and a = 10. Solve the triangle with a = 5, b = 6, and c = 7. Solve the triangle with a = 20, b = 15, and γ = 40◦. Tom Lewis () Trigonometry, Part I Spring Semester 2011 22 / 24 Solving non-right triangles Solution We can easily see that α = 89◦. By the law of sines, we can find the other two sides: b ≈ 5:45 and c ≈ 8:48. By the law of cosines, we find that cos(γ) = :2; hence, γ = 78:46◦. We can now use the law of sines to find the remaining angles: α ≈ 44:41◦ and β = 57:12◦. By the law of cosines we see that c ≈ 12:86. By the law of sines β ≈ 48:57◦; thus, α = 91:43◦. Try using the law of sines to find the measure of α first. What went wrong? Tom Lewis () Trigonometry, Part I Spring Semester 2011 23 / 24 Solving non-right triangles Problem From a certain point in his yard, a man is 100 feet from a hickory tree and 80 feet from an oak tree. What is the distance between these trees if the angle between their lines of sight is 73◦? Solution We can solve this by the law of cosines, obtaining d ≈ 108:27 ft. Tom Lewis () Trigonometry, Part I Spring Semester 2011 24 / 24.