1. Law of Sines

2. Introduction
In this section, we will solve 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. Remember ……

4. 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.

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

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

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

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

9. Law of Sines A b c C B a

10. Given Two Angles and One Side – AAS
For the triangle below C = 102, B = 29, and b = 28 feet. Find the remaining angle and sides.

11. 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 .

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

13. Example ASA A B C c a b

14. Example
A 46-foot telephone pole tilted at an angle of from the vertical casts a shadow on the ground. Find the length of the shadow to the nearest foot when the angle of elevation to the sun is
Draw a diagram Draw Then find the

15. Example cont
Since you know the measures of two angles of the triangle, and the length of a side opposite one of the angles you can use the Law of Sines to find the length of the shadow.
Law of Sines
Cross products
Divide each side by sin
Use a calculator.
Answer: The length of the shadow is about 75.9 feet.

16. Example
A 5-foot fishing pole is anchored to the edge of a dock. If the distance from the foot of the pole to the point where the fishing line meets the water is 45 feet, about how much fishing line that is cast out is above the surface of the water?
Answer: About 42 feet of the fishing line that is cast out is above the surface of the water.

17. The Ambiguous Case (SSA)

18. The Ambiguous Case (SSA)
In our first example we saw that two angles and one side determine a unique triangle.
However, if two sides and one opposite angle are given, three possible situations can occur:
(1) no triangle exists,
(2) one triangle exists, or
(3) two distinct triangles may satisfy the conditions.

19. Flowchart for ambiguous case

20. Steps for ambiguous case
Set up law of sines and solve for sin of missing angle
Use the flow chart

21. Example
In triangle ABC, b = 2, c = 8, and B = 120°. Solve the triangle. C B A? a 2 8
120°

22. Example
In triangle ABC, b = 2, c = 8, and B = 120°. Solve the triangle. C B A? a 2 8
120°

23. Example
In triangle ABC, c = 10, a = 6, and A = 28°. Solve the triangle.

24. Example
In triangle ABC, a = 6.5, b = 7.4, and B = 79°. Solve the triangle.

25. Area of an Oblique Triangle

26. 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)

27. 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.

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