2. Chapter 10: Applications of Trigonometry and Vectors
10.1 The Law of Sines
10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex Numbers
10.5 Powers and Roots of Complex Numbers
10.6 Polar Equations and Graphs
10.7 More Parametric Equations

3. 10.1 The Law of Sines Congruence Axioms
Side-Angle-Side (SAS) If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent.
Angle-Side-Angle (ASA) If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent.
Side-Side-Side (SSS) If three sides of one triangle are equal to three sides of a second triangle, the triangles are congruent.

4. 10.1 Data Required for Solving Oblique Triangles
Case 1 One side and two angles are known
(SAA or ASA).
Case 2 Two sides and one angle not included
between the two sides are known (SSA). This case may lead to zero, one, or two triangles.
Case 3 Two sides and the angle included between
the two sides are known
(SAS).
Case 4 Three sides are known
(SSS).

5. 10.1 Derivation of the Law of Sines
Start with an acute or obtuse triangle
and construct the perpendicular from
B to side AC. Let h be the height of this
perpendicular. Then c and a are the
hypotenuses of right triangle ADB
and BDC, respectively.

6. 10.1 The Law of Sines
In a similar way, by constructing perpendiculars from
other vertices, the following theorem can be proven.
Alternative forms are sometimes convenient to use:
Law of Sines
In any triangle ABC, with sides a, b, and c,

7. 10.1 Using the Law of Sines to Solve a Triangle
Example Solve triangle ABC if A = 32.0°, B = 81.8°,
and a = 42.9 centimeters.
Solution Draw the triangle
and label the known values.
Because A, B, and a are known,
we can apply the law of sines involving these variables.

8. 10.1 Using the Law of Sines to Solve a Triangle
To find C, use the fact that there are 180° in a triangle.
Now we can find c.

9. 10.1 Using the Law of Sines in an Application (ASA)
Example Two stations are on an east-west
line 110 miles apart. A forest fire is located
on a bearing of N 42° E from the western
station at A and a bearing of N 15° E from
the eastern station at B. How far is the fire
from the western station?
Solution Angle BAC = 90° – 42° = 48°
Angle B = 90° + 15° = 105°
Angle C = 180° – 105° – 48° = 27°
Using the law of sines to find b gives

10. Applying the Law of Sines
10.1 Ambiguous Case
Applying the Law of Sines
For any angle , –1  sin   1, if sin  = 1, then  = 90° and the triangle is a right triangle.
sin  = sin(180° –  ). (Supplementary angles have the same sine value.)
The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming that the triangle has sides that are all of different lengths).

11. 10.1 Ambiguous Case

12. 10.1 Ambiguous Case for Obtuse Angle A

13. 10.1 Solving the Ambiguous Case: No Such Triangle
Example Solve the triangle ABC if B = 55°40´,
b = 8.94 meters, and a = 25.1 meters.
Solution Use the law of sines to find A.
Since sin A cannot be greater than 1, the triangle does
not exist.

14. 10.1 Solving the Ambiguous Case: Two Triangles
Example
Solve the triangle ABC if A = 55.3°,
a = 22.8 feet, and b = 24.9 feet.
Solution

15. 10.1 Solving the Ambiguous Case: Two Triangles
To see if B2 = 116.1° is a valid possibility, add 116.1°
to the measure of A: 116.1° ° = 171.4°. Since
this sum is less than 180°, it is a valid triangle.
Now separate the triangles into two: AB1C1 and AB2C2.

16. 10.1 Solving the Ambiguous Case: Two Triangles
Now solve for triangle AB2C2.

17. 10.1 Number of Triangles Satisfying the Ambiguous Case (SSA)
Let sides a and b and angle A be given in triangle ABC. (The
law of sines can be used to calculate sin B.)
If sin B > 1, then no triangle satisfies the given conditions.
If sin B = 1, then one triangle satisfies the given conditions and B = 90°.
If 0 < sin B < 1, then either one or two triangles satisfy the given conditions
If sin B = k, then let B1 = sin-1 k and use B1 for B in the first triangle.
Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle.

18. 10.1 Solving the Ambiguous Case: One Triangle
Example Solve the triangle ABC, given A = 43.5°,
a = 10.7 inches, and c = 7.2 inches.
Solution
The other possible value for C:
C = 180° – 27.6° = 152.4°.
Add this to A: 152.4° ° = 195.9° > 180°
Therefore, there can be only one triangle.

19. 10.1 Solving the Ambiguous Case: One Triangle