1. Warm up
  A 5.2 m ladder leans against a wall. The bottom of the ladder is 1.9 m from the wall. What angle does the ladder make with the ground (to the nearest degree)?
Cos-1(1.9/5.2) = 69o

2. Objective: To use the Law of Sines in order to solve oblique triangles

3. Consider the first category, an acute triangle (, ,  are acute).

4. Create an altitude, h.

5. Theorem Law of Sines

6. The Law of Sines is used when we know any two angles and one side or when we know two sides and an angle opposite one of those sides.

7. Applying the Law of Sines
The Law of Sines may be used when the known parts of the triangle are:
1. one side and two angles (SAA), (ASA)
2. two sides and an angle opposite one of the sides (SSA)

8. Example:
In triangle ABC, angle A = 106 o, angle B = 31o and side a = 10 cm. Solve the triangle ABC by finding angle C and sides b and c.(round answers to 1 decimal place).
C= 43 degrees
c = 7.1 cm

9. Solution
Use the fact that the sum of all three angles of a triangle is equal to 180 o to write an equation in C. A + B + C = 180 o Solve for C. C = 180 o - (A + B) = 43 o

10. Solution (cont’d)
Use sine law to write an equation in b. a / sin(A) = b / sin(B) Solve for b. b = a sin (B) / sin(A) = (approximately) 5.4 cm
Use the sine law to write an equation in c. a / sin(A) = c / sin(C) Solve for c. c = a sin (C) / sin(A) = (approximately) 7.1 cm

11. Area of an Oblique Triangle
The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle.
Referring to the triangles below, that each triangle has a height of h = b sin A.
A is acute.
A is obtuse.

12. Area of a Triangle - SAS
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is ½ base ● height
Base = b
Height = c ● sin A
 Area K= ½ bc(sinA) A B C c a b h
Looking at this from all three sides:
K = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

13. Example: Find the area of
given a = 32 m, b = 9 m, and

14. Area of a Triangle
You can also find the area if you know one side and 2 angles based on the Law of Sines.
so, substitute for b in the last
equation, K = ½ bc(sinA) gives you

15. Area of a Triangle
Find the area of triangle JKL if j=45.7, K=111.1o, and L=27.3o.
673.0 units sq.

16. Law of Sines practice
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17. Warm up
Using the triangle above, A = 50o, B = 65o and a = 12.  Solve the triangle.
/ C = 65o b = 14.2 c = 14.2  

18. Lesson 5-7 Law of Sines the Ambiguous Case
Objective: To determine whether a triangle has zero, one or two solutions and solve using the Law of Sines.

19. The Ambiguous Case – SSA
In this case, you may have information that results in one triangle, two triangles, or no triangles.

20. SSA – No Solution
Two sides and an angle opposite one of the sides.

21. By the law of sines,

22. Therefore, there is no value for  that exists! No Solution!
Thus,
Therefore, there is no value for  that exists! No Solution!

23. SSA – Two Solutions

24. By the law of sines,

25. So that,

26. Case Case 2
Both triangles are valid! Therefore, we have two solutions.

27. Case 1

28. Case 2

29. For SSA Triangles: If A < 90° If A ≥ 90° a < b a ≥ b 1 Solution
1. a < b(sin A) No Solution
a = b(sin A) 1 Solution
a > b(sin A) 2 Solution
a ≥ b 1 Solution
If A ≥ 90°
a ≤ b No Solution
a > b 1 Solution

30. Practice
Solve the triangle: A = 42°, a = 11, and b = 6