The Law of Cosines

Key Questions

  • When can the law of cosines be used?

    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.

    Massimiliano · 1 · Jan 29 2015
  • Does the law of cosines work for any triangle?

    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.

    Zor Shekhtman · 1 · Dec 16 2014
  • How do you find the angle in the law of cosines?

    Answer:

    cos(gamma)=(a^2+b^2-c^2)/(2ab)

    Explanation:

    We have
    c^2=a^2+b^2-2abcos(gamma)
    so
    2abcos(gamma)=a^2+b^2-c^2
    Isolating cos(gamma)
    cos(gamma)=(a^2+b^2-c^2)/(2ab)

    Sonnhard · 1 · May 27 2018
  • What is the law of cosines?

    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

    Gió · 1 · Dec 21 2014

Questions

Triangles and Vectors