Chapter 2 Trigonometry 2.3 The Sine Law Pre-Calculus 11

Recall 30° 0.174 Trig Equations Angle or ratio? Angle or ratio? Angle or ratio? Remember that the ratio is the ratio between side lengths of a triangle. The particular side lengths involved in the ratio are determined by the trig function Pre-Calculus 11

? Solving Non-Right Triangles with Primary Trig Ratios Stan observes that the angle of elevation of a plane to be 510. At the same time, Paul observes it to be 340. The pilot used a rangefinder to determine that Stan is 190.85m from the plane and Paul is 265.24m from the plane. How far apart are Stan and Paul? (nearest m) 190.85 m 265.24 m Paul Stan 510 340 ? Pre-Calculus 11

? Solving with Primary Trig Ratios Method 1: draw a straight line thought the middle of the triangle to create two right triangles. Use the trig ratios to solve for the length of the side Solve for BD and CD to find the total distance 510 340 A B C Stan Paul 265.24 m ? 190.85 m Total Distance = BD + CD sohcahtoa D Stan and Paul are 340 m apart Alternate method: Solve non-Right Triangle Pre-Calculus 11

How can we expand this formula to include angle A and side a? An oblique triangle is a triangle that does not contain a right angle. We can solve these triangles using a different method Using the definition of sine ratio: A B C b c a h Isolate the variable h: D Using the Transitive Property: Divide by bc Or Divide by sinBsinC How can we expand this formula to include angle A and side a? Pre-Calculus 11

The Law of Sines For an oblique triangles or right triangles, when you are given SSA (Side-Side-Angle) or ASA (Angle-Side-Angle): a b c B C A capital letter (A, B, C) is an angle lower case letter (a, b, c) is the side length opposite that angle Pre-Calculus 11

Proving Equivalence What do you notice about each ratio?   1 60° 30° 2 What do you notice about each ratio? They are all equal Would equivalence change if the reciprocals were written? No. They would still all be equal ratios Pre-Calculus 11

? or Solving with Primary Trig Ratios Method 2: Solve non-Right Triangle using Sine Law When we are given either: two angles and one side (ASA) or two sides and an angle opposite one of the given sides (SSA) we can use the Sine Law 510 340 A B C Stan Paul 265.24 m ? 190.85 m 950 We need to find angle A. Recall that the sum of the angles in any triangle equals 180° so ∠A = 180° - ∠B - ∠C → ∠A = 180° - 51° - 34° ∠A = 95° answer is the same How would the solution be affected if you would have chosen or Stan and Paul are 340 m apart Pre-Calculus 11

Applying the Law of Sines (nearest 100th for lengths, 10th for angles) Examples 600 c Determine the length of side c Determine angle 𝜃: Find ㄥC : 1800- (750 + 450) = 600 c = 12.25 cm Pre-Calculus 11

The Sine Law State the formula for the Law of Sines. What specific information must be given in a triangle to apply the Law of Sines? Given: AAS or SSA Given: two angles and one side or two sides and an angle opposite one of the given sides What information must you have to define the specific ratio? one side length and its opposite angle

Sketching a Diagram Chapter A ship at sea is sighted from two coast guard observation posts on shore. The angle between the line of sight from post A and the line between the two posts is 110o. From post B, the angle is 30o. If the two observation posts are 15 km apart, sketch a diagram that could be used to determine the distance, to the nearest tenth of a kilometre, between each observation post and the ship. 2 110° 30° coast guard observation post A coast guard observation post B 15 km AAS Write the equations that could be used to determine the distances to the observation posts from the ship. Pre-Calculus 11

2.3 B The Ambiguous Case of the Sine Law Chapter 2 Trigonometry 2.3 B The Ambiguous Case of the Sine Law Acute ∠A Pre-Calculus 11

2.3 B The Ambiguous Case of the Sine Law Chapter 2 Trigonometry 2.3 B The Ambiguous Case of the Sine Law Obtuse ∠A Pre-Calculus 11

SSA Determine the Number of Triangles When you solve for the measure of angle B, there may be two solutions. Angle B could be in QI or QII. When two sides and the non-included angle (SSA) of a triangle are given, the triangle may not be unique. It is possible that no triangle, one triangle, or two triangles exist with the given measurements. This is referred to as the Ambiguous Case of the Law of Sines. Pre-Calculus 11

Angle A is Obtuse (> 90°) Ambiguous Case Angle A is Obtuse (> 90°) You must compare the length of side a to side b. Obtuse - has an angle > 90o Pre-Calculus 11

Angle A is Acute(< 90°) Ambiguous Case Angle A is Acute(< 90°) You must compare the length of side a to side b and also the height of the triangle, h. Number of Triangles Sketch Conditions Acute - all three angles are < 90o Pre-Calculus 11

Determine the Number of Triangles SSA When two sides and the non-included angle of a triangle are given, the triangle may not be unique. It is possible that no triangle, one triangle, or two triangles exist with the given measurements. C If bsinA < a < b, (bsinA = h) then 2 triangles are possible. AB1C and ABC a b a h = bsinA A B1 B ㄥC is also different in each triangle This is referred to as the Ambiguous Case of SSA. Pre-Calculus 11

Exploring the relationship between ∠B’ and ∠B (ambiguous case) Consider the following ambiguous case: Determine ㄥx : In ΔB’BC, we have an isosceles triangle since the legs of the triangle have identical length, a, we know that ㄥB and ㄥx are equal (ㄥB = ㄥx) via the base angles theorem x Determine ㄥB’: Notice that ㄥx and ㄥB’ are supplementary angles. So ㄥB’ = 180° - ㄥx (or ㄥB). ㄥB is a quadrant I angle (and reference angle 𝞱R) ㄥB’ is a quadrant II angle (180° - 𝞱R) Pre-Calculus 11

Therefore, 2 triangles exist. Solve The Ambiguous Case Triangle - Method 1 - check conditions First check: a < b TRUE Then calculate bsinA (height) bsinA = 15sin40 = 9.642 9.642 < 10 < 15 bsinA< a < b h < a < b Therefore, 2 triangles exist. C C 15 15 10 10 400 400 In the second case angle B is now a second quadrant angle. Therefore, A B A B1 Pre-Calculus 11

Solve The Ambiguous Case Triangle - Method 2 - no conditions Use sine law to calculate ㄥB: Determine ㄥB1 in QII: If the sum of given angle (ㄥA) and ㄥB1 is less than 180° then there is 2 triangles: Therefore, 2 triangles exist

How Many Solutions are Possible (SSA)? C 17 20 h Check: Angle B acute or obtuse? 530 B A1 A Check: Is side opposite the given angle < other given side? 17 < 20 Check: Is height < opposite side < hypotenuse side? 20sin53º < 17 < 20 15.97 < 17 < 20 Conclusion: Ambiguous Case: two triangles are possible. Pre-Calculus 11

Assignment How Many Solutions are Possible (SSA)? Suggested Questions Check: Angle F acute or obtuse? Check: Is side opposite the given angle < other given side? 24 cm < 18 cm Conclusion: Only one triangle is possible Assignment Suggested Questions Page 108: (1-6)ac, 8a, 10-13, 15, 17, 18, 24 Pre-Calculus 11