1. Chapter 2 Trigonometry
2.3 The Sine Law
Pre-Calculus 11

2. 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

3. ? 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 The pilot used a rangefinder to determine that Stan is m from the plane and Paul is m from the plane. How far apart are Stan and Paul?
(nearest m) m m
Paul
Stan
510
340 ?
Pre-Calculus 11

4. ? 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 m ? m
Total Distance = BD + CD
sohcahtoa D
Stan and Paul are 340 m apart
Alternate method:
Solve non-Right Triangle
Pre-Calculus 11

5. 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

6. 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

7. 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

8. ? 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 m ? 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

9. 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- ( ) = 600
c = cm
Pre-Calculus 11

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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