1. 13. Law of Sines

2. Introduction In this section, we will solve (find all
the sides and angles of) oblique triangles – triangles that have no right angles.
As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c.
To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.

3. 4 cases Two angles and any side are known (AAS or ASA)
Two sides and an angle opposite one of them are known (SSA)
Three sides are known (SSS)
Two sides and their included angle are known (SAS)
The first two cases can be solved using the Law of Sines, the last two cases require the Law of Cosines.

4. A S A
ASA S A A
SAA
CASE 1: ASA or SAA Law of sines

5. S A S
CASE 2: SSA Law of sines

6. S A S
CASE 3: SAS Law of cosines

7. S S S
CASE 4: SSS Law of cosines

8. Deriving the Law of Sines
Triangle ABC does not have a right angle so we drop a perpendicular from
Angle B
We call this h and now we have a right angle so we can use SOHCAHTOA B c a h A C b
sin A=h/c, h = csinA
sin C=h/a, h = asinC
csinA=asinC

9. Law of Sines A b c C B a

10. Remember ……

11. In a triangleC = 102, B = 29, and b = 28 feet. Solve the triangle
Case 1 - AAS
In a triangleC = 102, B = 29, and b = 28 feet. Solve the triangle

12. Example AAS - Solution The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102
= 49.
By the Law of Sines, you have .

13. Example AAS – Solution
cont’d
Using b = 28 produces
and

14. Case 1 - ASA A B C c a b
C=70o
b=44.1
a=32.7

15. Area of an Oblique Triangle (Given SAS)

16. Area of a Triangle Area = ½ bh sin A = h/c h = c sin A
Area = ½ b c sin A

17. Area of a Triangle Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A) c Ah
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

18. Example – Finding the Area of a Triangular Lot
Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102.
Solution:
Consider a = 90 meters, b = 52 meters, and the included angle C = 102
Then, the area of the triangle is
Area = ½ ab sin C
= ½ (90)(52)(sin102)
 2289 square meters.

19. Try It Out Determine the area of these triangles 42.8° 127° 76.3° 17.924 12
76.3°