The Law of Cosines

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Using the Sine and Cosine Rule To Solve Unknown Angles

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Key Questions

  • Cosider the triangle:

    enter image source here
    (Picture source: Wikipedia)

    you can relate the sides of this triangle in a kind of "extended" form of Pitagora's Theorem giving:

    #a^2=b^2+c^2-2bc*cos(alpha)#
    #b^2=a^2+c^2-2ac*cos(beta)#
    #c^2=a^2+b^2-2ab*cos(gamma)#

    As you can see you use this law when your triangle is not a right-angled one.

    Example:
    Consider the above triangle in which:
    #a=8 cm#
    #c=10 cm#
    #beta=60°# therefore:
    #b^2=a^2+c^2-2ac*cos(beta)#
    #b^2=8^2+10^2-2*8*10*cos(60°)# but #cos(60°)=1/2#
    so: #b^2=84 and b=sqrt(84)= 9,2 cm#

  • The Law of Cosine can only be used if two sides and their enclosed angle are known.

    In all other cases the Law of Sines has to be used.

  • Yes, the Law of Cosines works for all triangles.

    However, the proof depends on the shape a triangle, more precisely, how an altitude from some vertex falls onto the opposite side.

    For example, consider a triangle #Delta ABC# with vertices #A#, #B# and #C#, corresponding angles #alpha#, #beta# and #gamma# and correspondingly opposite sides #a#, #b# and #c#.
    Let's prove the Law of Cosines that states:
    #a^2+b^2-2*a*b*cos(gamma) = c^2#

    Let's draw altitude #AH# from vertex #A# to an opposite side #BC# with an intersection of this altitude and a side #BC# at point #H#.
    There are different cases of a location of point #H# relatively to vertices #B# and #C#.
    It can lie in between vertices #B# and #C#.
    It can lie outside of #BC# on a continuation of this side beyond vertex #B# or beyond vertex #C#.

    Assume that a base of this altitude, point #H#, is lying on the continuation of #BC# beyond a point #C# (so, #C# is in between #B# and #H#) and prove the Law of Cosines in this case. Other cases are similar to this one.

    Let's use the following symbols for segments involved:
    #AH# is #h#
    #BH# is #a_1#
    #CH# is #a_2#
    Then, since #C# lies in between #B# and #H#,
    #a = a_1 - a_2# or #a_1=a+a_2#
    Since both #Delta ABH# and #Delta ACH# are right triangles, by trigonometric dependency between hypotenuse, catheti and angles and by Pythagorean Theorem
    #h = b*sin(pi-gamma) = b*sin(gamma)#
    #a_2 = b*cos(pi-gamma) = -b*cos(gamma)#
    #c^2 = h^2+a_1^2 = b^2*sin^2(gamma)+(a-b*cos(y))^2 =#
    #=b^2*sin^2(gamma)+a^2-2*a*b*cos(gamma)+b^2*cos^2(gamma)=#
    #=a^2+b^2(sin^2(gamma)+cos^2(gamma))-2*a*b*cos(gamma)=#
    #=a^2+b^2-2*a*b*cos(gamma)#

    End of Proof

    When point #H# lies in between vertices #B# and #C# or on a continuation of side #BC# beyond vertex B, the proof is similar.

    See Unizor Trigonometry - Simple Identities - Law of Cosines for visual presentation and more detailed information.

  • Find C using the Law of Cosines.

    #c^2=b^2+a^2-2abCos(C)#
    This can also be written as:
    #a^2=b^2+c^2-2bcCosA#
    #b^2=a^2+c^2-2acCosB#

    Say you have a triangle with Angles A, B, and C and sides a, b, and c as shown below.
    http://2000clicks.com/mathhelp/GeometryCosinesFig1.gif
    In order to find the angle, the problem will most likely give you the length of the 3 sides.

    Just for explanation purposes, let's give these side lengths values.
    a=9, c=8, and b=5.

    If we substituted all of these values into our Law of Cosines equation, the new equation would look like:
    #8^2=5^2+9^2-2(5)(9)Cos(C)#

    You wold then simplify the equation so that it comes down to
    -42=-90Cos(C)

    Isolate the Cos(C):
    Cos(C)=42/90

    You now have to take the inverse of Cosine to isolate Angle C.
    #C=Cos^-1(42/90)#
    Round to however your teacher wants you to, but you should get about 62.18 degrees.

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