Trigonometry CHAPTER 2

Chapter 2: Trigonometry 2.1 – ANGLES IN STANDARD POSITION

Warm Up Test

STANDARD POSITION

Which diagrams shows an angle of 70° at standard position? How might we define an angle at standard position? An angle in standard position is the position of an angle when its initial arm is on the positive x-axis and its vertex is at the origin.

 Fold the paper that was handed out in half lengthwise and make a crease down the middle.  Take corner C to the centre fold line and make a crease, DE.  Fold corner B so that BE lies on the edge of segment DE. The fold will be along line segment C’E. Fold the overlap (the grey-shaded region) under to complete the equilateral triangle. ACTIVITY

 Assume that our Equilateral Triangle has side lengths of 2 units.  To obtain a 30°-60°-90° triangle, fold the triangle in half.  Label the angles in the triangle as 30°, 60° and 90°.  Use the Pythagorean Theorem to determine the exact measure of the third side of the triangle. ACTIVITY (CONT) c 2 = a 2 + b 2 c 2 =  c 2 = = 5  c =

QUADRANTS

A reference angle is the acute angle whose vertex is the origin and whose arms are the terminal arm of the angle and the x- axis. REFERENCE ANGLE

 For angles of 30°, 45°, and 60°, we can determine the exact values of trigonometric ratios. SPECIAL TRIANGLES What length is c?

sin(45°) = cos(45°) = tan(45°) = sin(30°) = sin(60°) = cos(30°) = cos(60°) = tan(30°) = tan(60°) =

EXAMPLE Determine the reference angle, θ R, if θ = 300°.

EXAMPLE Allie is learning to play the piano. Her teacher uses a metronome to help her keep time. The pendulum arm of the metronome is 10 cm long. For one particular tempo, the setting results in the arm moving back and forth from a start position of 60° to 120°. What horizontal distance does the tip of the arm move in one beat? Give an exact answer. Draw a diagram: I know the hypotenuse and the angle, and want the adjacent, so I should use cosine. cos(60) = a/10  a = 10cos(60) = 10(1/2) (from the special triangle)  a = 5 So, half-a-beat is 5 cm, so one whole beat must travel a horizontal distance of 10 cm.

Independent Practice PG # 3, 5, 8, 10, 12, 14, 16, 18, 19, 21, 23, 24

Chapter 2: Trigonometry 2.2 – TRIGONOMETRIC RATIOS OF ANY ANGLE

 Get out some graph paper. Plot the point A(3,4). Which quadrant is it in? Draw the angle in standard position with terminal arm going through point A.  Draw a line going down from point A to the x-axis.  What is the length of r? How could we find out?  r 2 = = 25  r = 5  What are the trig ratios for θ?  sinθ = 4/5  cosθ = 3/5  tanθ = 4/3  What is the measure of θ?  θ = sin -1 (4/5) = 53° TRIGONOMETRIC RATIOS FOR ANGLES GREATER THAN 90°

 What is the relationship of each trig ratio and r and A’s coordinates?  sinθ = y/r, cosθ = x/r, tanθ = y/x  Reflect point A in the y-axis and call it point C. Draw a line segment to point C. What are its coordinates? Draw a line down from point C to the x-axis.  Label the point of intersection with the x-axis as point D.  Using our relationships of the trig ratios and C’s coordinates and r, what are the the trig ratios of ∠ COB.  sin ∠ COB = y/r = 4/5  cos ∠ COB = x/r = -3/5  tan ∠ COB = y/x = -4/3  ∠ COB = cos -1 (-3/5) = 127°  ∠ COD = 180 – 127 = 53° TRIG RATIOS CONTINUED

TRIG RATIOS ∠ AOB sin ∠ AOB = 4/5 cos ∠ AOB = 3/5 tan ∠ AOB = 4/3 ∠ COB sin ∠ COB = 4/5 cos ∠ COB = -3/5 tan ∠ COB = -4/3 Why are some of the trig ratios positive while others are negative? What do you think would happen if we reflected point C in the x-axis, to obtain point E? What would its coordinates be? Which trig ratios would be positive and which would be negative? C(-3, 4) E(-3, -4) sin ∠ EOB = -3/5 cos ∠ EOB = -4/5 tan ∠ EOB = 4/3

TRIG RATIOS C(-3, 4) E(-3, -4) What about point G(3,-4)? sin ∠ GOB = -4/5 cos ∠ GOB = 3/5 tan ∠ GOB = -4/3 G(3,-4) So, recall: ∠ AOB (Quadrant I) sin ∠ AOB = 4/5 cos ∠ AOB = 3/5 tan ∠ AOB = 4/3 ∠ COB (Quadrant II) sin ∠ COB = 4/5 cos ∠ COB = -3/5 tan ∠ COB = -4/3 ∠ EOB (Quadrant III) sin ∠ EOB = -4/5 cos ∠ EOB = -3/5 tan ∠ EOB = 4/3 ∠ GOB (Quadrant IV) sin ∠ GOB = -4/5 cos ∠ GOB = 3/5 tan ∠ GOB = -4/3

∠ AOB (Quadrant I) sin ∠ AOB = 4/5 cos ∠ AOB = 3/5 tan ∠ AOB = 4/3 ∠ COB (Quadrant II) sin ∠ COB = 4/5 cos ∠ COB = -3/5 tan ∠ COB = -4/3 ∠ EOB (Quadrant III) sin ∠ EOB = -4/5 cos ∠ EOB = -3/5 tan ∠ EOB = 4/3 ∠ GOB (Quadrant IV) sin ∠ GOB = -4/5 cos ∠ GOB = 3/5 tan ∠ GOB = -4/3 Which ratio is positive in each quadrant? All Sine Tangent Cosine

REVIEW For any angle 0° < θ < 180°, formed with its terminal arm through any point P(x,y), the trigonometric ratios can be defined as: sinθ = y/rcosθ = x/rtanθ = y/x All Sine Tangent Cosine To help remember which ratio is positive in each quadrant, remember “CAST”:

EXAMPLE Determine the exact value of cos135°. What’s the reference angle? What do we know about a triangle with an angle of 45°? (hint: special triangles) We know the side lengths! So, what’s the rule for cosine? cos135 = x/r We know it should be negative, because quadrant II is only positive for sine!

TRY IT! Find the exact value of sin240°.

EXAMPLE Suppose θ is an angle in standard position with terminal arm in quadrant III, and cosθ = -3/4. What are the exact values of sinθ and tanθ? Draw a diagram: cosθ = x/r (from our rules!)  Since it’s in quadrant III, x = -3, and r = 4. What’s y? r 2 = x 2 + y 2  4 2 = (-3) 2 + y 2  y 2 = 16 – 9 = 7  Why is y negative? Now we can compute sine and tangent: sinθ = y/r  tanθ = y/x 

EXAMPLE Solve for θ: Since θ is between 0° and 180°, it must lie in either quadrant I or II. Cosine is negative, so which quadrant must it be in?  All trig ratios are positive in quadrant I, so it must be in quadrant II! Remember the special triangle: So the reference angle must be 30° 180 = 30 + θ  θ = 180 – 30  θ = 150°

Independent Practice P #3, 5, 6, 9, 13, 15, 17, 18, 21, 23, 25, 27, 28

Chapter 2: Trigonometry 2.3 – THE SINE LAW

HANDOUT Complete the handout on the Sine Law to the best of your abilities. Make sure to answer all the questions in their entirety as this is a summative assessment.

OR

EXAMPLE In ΔPQR, ∠ P = 36°, p = 24.8 m, and q = 23.4 m. Determine the measure of ∠ R, to the nearest degree, and side length r to the nearest tenth of a metre. Draw a diagram: ∠ P = 36° p = 24.8 m ∠ Q = ? q = 23.4 m ∠ R = ? r = ? Since we don’t have either the side or the angle for R, we need to figure one of them out to use the sine law. Which can we work out? What if we solve for ∠ Q first? Now, if we know ∠ Q and ∠ P, how can we find ∠ R? So, to the nearest degree, ∠ R = 110°, and to the nearest metre, r = 39.6 m.

 When solving a triangle, sometimes a solution doesn’t actually exist—we call this the ambiguous case. If you are given ASA (two angles, with a side between them), then the triangle is uniquely defined. However, if you are given SSA, the ambiguous case may occur. In this case, there are three possible outcomes:  No triangle exists that has the given measures; there is no solution.  One triangle exists that has the given measure; there is one solution  Two distinct triangles exist that have the given measures; there are two distinct solutions THE AMBIGUOUS CASE

Suppose you are given the measures of side b and ∠ A of ΔABC. You can find the height of the triangle by using h = bsinA. In ΔABC, ∠ A and side b are constant because they are given. Consider the different possible lengths of side af or a triangle where ∠ A is acute:

Suppose you are given the measures of side b and ∠ A of ΔABC. You can find the height of the triangle by using h = bsinA. In ΔABC, ∠ A and side b are constant because they are given. Consider the different possible lengths of side a for a triangle where ∠ A is acute:

When ∠ A is obtuse:

IN SUMMARY

EXAMPLE In ΔABC, ∠ A = 30°, a = 24 cm, and b = 42 cm. Determine the measures of the other sides and angles. Round your answers to the nearest unit. ∠ A = 30a = 24 cm ∠ B = ?b = 42 cm ∠ C = ?c = ? We have SSA, so it’s the ambiguous case. ∠ A is acute, so check the conditions: a < bsinA : no solution a = bsinA : one solution a > bsinA : two solutions Possible diagram: sin30 = h/42  h = 42sin30  h = > 21, So a > bsinA So there are two solutions. But remember, there are TWO possible solutions. ∠ B being 61° is only one of them. What’s the other?

EXAMPLE (CONTINUED) ∠ A = 30a = 24 cm ∠ B = ?b = 42 cm ∠ C = ?c = ? So, ∠ B = 61°  ∠ C = 180 – 61 – 30  ∠ C= 89° Case 1: Case 2: In case 2, use ∠ B = 61° as the reference angle in quadrant II.  ∠ B = 180 – 61 = 119°  ∠ C = 180 – 119 – 30 = 31° Either: ∠ B = 61°, ∠ C = 89° and c = 48 cm OR ∠ B = 119°, ∠ C = 31° and c =25 cm

Independent practice P # 4, 5, 8, 10, 12, 14, 16, 18, 20, 22

Chapter 2: Trigonometry 2.4 – THE COSINE LAW

THE COSINE LAW Everyone draw an acute triangle ΔABC. Measure the sides, and write down the measurements. Make sure your sides and vertices are labeled. We’ll now fill out the handed-out chart together: Triangle Side Lengths (cm)c2c2 a 2 + b 2 2abcosC a = 3, b = 4, c = 5

THE COSINE LAW The cosine law describes the relationship between the cosine of an angle and the lengths of the three sides of any triangle. For any ΔABC, where a, b, and c are the lengths of the sides opposite to ∠ A, ∠ B, and ∠ C, respectively, the cosine law states that c 2 = a 2 + b 2 – 2ab cosC. You can also express the formula in different forms to find the lengths of the other sides of the triangle: c 2 = a 2 + b 2 – 2ab cosC a 2 = b 2 + c 2 – 2bc cosA b 2 = a 2 + c 2 – 2ac cosB

PROOF OF THE COSINE LAW In any acute triangle, a 2 = b 2 + c 2 – 2bc cosA b 2 = a 2 + c 2 – 2ac cosB c 2 = a 2 + b 2 – 2ab cosC Proof: h 2 = c 2 – x 2 h 2 = b 2 – y 2  c 2 – x 2 = b 2 – y 2  c 2 = x 2 + b 2 – y 2  c 2 = (a – y) 2 + b 2 – y 2 (since x = a – y)  c 2 = a 2 – 2ay + y 2 + b 2 – y 2  c 2 = a 2 + b 2 – 2ay cosC = y/b(cos = adj/hyp)  b cosC = y  c 2 = a 2 + b 2 – 2ay  c 2 = a 2 + b 2 – 2ab cosC The other parts can be proven similarly.

EXAMPLE A surveyor needs to find the length of a swampy area near Fishing Lake, Manitoba. The surveyor sets up her transit at a point A. She measures the distance to one end of the swamp as m, the distance to the opposite end of the swamp as m, and the angle of sight between the two as 78.6°. Determine the length of the swampy area, to the nearest tenth of a metre. Use the cosine law: a 2 = b 2 + c 2 – 2bc cosA  a 2 = – 2(692.6)(468.2)cos(78.6°)  a 2 = …  a = …  The length of the swampy area if m, to the nearest tenth of a metre.

EXAMPLE In ΔABC, a = 11, b = 5, and ∠ C = 20°. Sketch a diagram and determine the length of the unknown side and the measures of the unknown angles, to the nearest tenth. Sketch a diagram: ∠ A = ?a = 11 ∠ B = ?b = 5 ∠ C = 20c = ? Use the cosine law: c 2 = a 2 + b 2 – 2ab cosC  c 2 = (11) 2 + (5) 2 – 2(11)(5) cos20  c 2 = …  c = 6.529… To solve for the angles we could use either the cosine or the sine law. For practice, we’ll use the cosine law: a 2 = b 2 + c 2 – 2bc cosA  cosA = (b 2 + c 2 – a 2 )/2bc  cosA = (5 2 + (6.529) 2 – 11 2 )/(2*5*6.529)  ∠ A = cos -1 [(5 2 + (6.529) 2 – 11 2 )/(2*5*6.529)]  ∠ A = ° ∠ C = 180 – – 20 = 15.2° So, c = 6.5, ∠ A = 144.8°, ∠ C = 15.2°.

Independent practice PG , # 3, 5, 8, 10, 13, 16, 18, 20, 22, 23, 25, 27, 28, 31