Law of Cosine

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extending Pythagorus

In the diagram below, we drop an altitude from the top corner (angle $C$). This altitude (length $h$) is perpendicular to the opposite side, forming two right triangles inside the acute triangle.

Points, Line Segments, and Lengths

If $A$ and $D$ are two points, then

the actual line segment itself connecting $A$ and $D$ is denoted as $\overline {AD}$.
the length of this line segment is denoted as $m(\overline {AD})$. $m$ stands for measurement.
If $A$ is a point, which is serving as the vertex of an angle, then the angle is denoted by $\measuredangle A$.

From the right triangle on the left in the diagram, we can see that $\cos (\measuredangle A) = \frac {m(\overline {AD})}{b}$ or $m(\overline {AD}) = b \cos (\measuredangle A)$.

Note: When it is clear, the angle sign is almost always dropped.

From the right triangle on the left in the diagram, we can see that $\cos (A) = \frac {m(\overline {AD})}{b}$ or $m(\overline {AD}) = b \cos (A)$.

Subtracting this from $m(\overline {AB})$, gives us $m(\overline {DB}) = c - b \cos (A)$.

From the right triangle on the left in the diagram, we can see that $\sin (A) = \frac {h}{b}$ or $h = b \sin (A)$.

From the right triangle on the right in the diagram, the Pythagorean Theorem gives us $a^2 = (b \sin (A))^2 + (c - b \cos (A))^2$.

Multiplying everything out gives

\[ a^2 = b^2 \sin ^2(A) + c^2 - 2 b c \cos (A) + b^2 \cos ^2(A) \]

\[ a^2 = b^2 \sin ^2(A) + b^2 \cos ^2(A) + c^2 - 2 b c \cos (A) \]

\[ a^2 = b^2 (\sin ^2(A) + \cos ^2(A)) + c^2 - 2 b c \cos (A) \]

\[ a^2 = b^2 + c^2 - 2 b c \cos (A) \]

Law of Cosines

For any trinagle with angles $A$, $B$, and $C$, and opposite sides $a$, $b$, and $c$, respectively,

\[ a^2 = b^2 + c^2 - 2 b c \cos (A) \]

\[ b^2 = a^2 + c^2 - 2 a c \cos (B) \]

\[ c^2 = a^2 + b^2 - 2 a b \cos (C) \]

Triangle

Suppose we have a triangle with $a=6$, $b=9$, and $c=11$. Approximate the measurement of angle $A$.

\[ 6^2 = 9^2 + 11^2 - 2 \cdot 9 \cdot 11 \cos (A) \]

\[ \frac {36 - 81 - 121}{- 2 \cdot 9 \cdot 11} = \cos (A) \]

\[ \answer [tolerance=0.001]{0.83838383} \approx \cos (A) \]

\[ A \approx \cos ^{-1}(0.83838383) \approx \answer [tolerance=0.01]{33.030}^{\circ } \]

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more examples can be found by following this link
More Examples of Triangles

2025-05-17 23:20:15