Trig. Equations

In trigonometry, there are a few, more complicated equations that allow us to solve more complex problems. These equations can help solve oblique triangles or triangles that have no right angles. The first law we will learn is the Law of Sines.

The Law of Sines states: If ABC is a triangle with sides a, b, c, then                                                                                 AAS, ASA, SSA, SSS, SAS

The Law of Sines can also be written in reciprocal form               

To solve an oblique triangle, the measure of at least one side and any two other parts of the triangle are needed such as the measure of the two other sides, two angles, or one angle and one side.

Example 1

A triangle with two angles and one side-AAS

C=102.3º

B=28.7º

b=27.7 ft.

Find the remaining angle and sides.

SolutionThe third angel of the triangle is        A=180º - B - C       =180º - 28.7 - 102.3º       = 49.0ºBy law of Sines, you have        __a__      __b__       ___c___        sin49º = sin28.7º  = sin102.3º                                                                           Using b = 27.4 produces                   ( _27.4_        a=       sin 28.7º)(sin 49º) ≈43.06 ft. and                     (__27.4_        c=         sin 28.7º)(sin 102.3º) ≈ 55.75 ft. Law of Cosines

    The two cases remaining that cannot be solved by sine are SSS, and SAS. To use the Law of Sines, you need to know at least one side and its opposite angle. If you are given three sides (SSS), or two sides and their included angle (SAS), none of the rations in the Law of Sines would be complete. In such cases you can use the law Law of Cosines.

                    Law of Cosines

        Standard Form                                Alternative Form

    a2 = b2+ c2 - 2bc cos A                      cos A =  b2+ c2- a2

                                                                                 2bc

    b2 = a2 + c2 - 2ac cos B                   cos B = a2 + c2 - b2

                                                                               2ac

    c2 = a2 + b2 - 2ab cos C                   cos C = a2 + b2 - c2

                                                                                 2ab

Three Sides of a Triangle - SSS

    Find the three angles of the triangle.

SolutionIts a good idea first to find the angle opposite the longest side - side b in this case. Using the Law if Cosines, you find that    cos B = a2 + c2 - b2   =   82 +142 - 192   ≈   -0.45089                           2ac                    2(8)(14)Because cos B is negative, you know that B is an obtuse angle given by B ≈ 116.80º. At this point knowing that B ≈ 116.80º, it is simpler to use the Law of Sines to determine A.     sin A = a (sin B   ≈  8 (sin 116.80º  ≈  0.37582                        b  )                19       )Because B is obtuse, you know that A must be acute, because a triangle can have at most one obtuse angle. So, A ≈ 22.08º and     C ≈ 180º - 22.08º - 116.80º = 41.12º