Law of Tangents/Corollary/Examples/Arbitrary Example 1

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Example of Use of Corollary to Law of Tangents

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Let $\triangle ABC$ be such that:

\(\ds a\) \(=\) \(\ds 18.4\)
\(\ds b\) \(=\) \(\ds 12.2\)
\(\ds C\) \(=\) \(\ds 42 \degrees\)

where the units of measurement of $a$ and $b$ are arbitrary.


Then:

\(\ds A\) \(=\) \(\ds 96 \degrees 50'\)
\(\ds B\) \(=\) \(\ds 41 \degrees 10'\)
\(\ds c\) \(=\) \(\ds 12.4\)


Proof

First:

\(\ds A + B\) \(=\) \(\ds 180 \degrees - C\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \) \(=\) \(\ds 180 \degrees - 42 \degrees\)
\(\ds \) \(=\) \(\ds 138 \degrees\)


Then:

\(\ds \tan \dfrac {A - B} 2\) \(=\) \(\ds \dfrac {18.4 - 12.2} {18.4 + 12.2} \cot 21 \degrees\) Corollary to Law of Tangents
\(\ds \) \(=\) \(\ds \dfrac {6.2} {30.6} \cot 21 \degrees\) Corollary to Law of Tangents
\(\ds \leadsto \ \ \) \(\ds \dfrac {A - B} 2\) \(=\) \(\ds 27 \degrees 50'\)
\(\ds \leadsto \ \ \) \(\ds A - B\) \(=\) \(\ds 55 \degrees 40'\)
\(\ds A + B\) \(=\) \(\ds 138 \degrees\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds 96 \degrees 50'\)
\(\ds B\) \(=\) \(\ds 41 \degrees 10'\)


Then we have:

\(\ds \dfrac c {\sin C}\) \(=\) \(\ds \dfrac b {\sin B}\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds \dfrac {12.2 \sin 42 \degrees} {\sin 41 \degrees 10'}\)
\(\ds \) \(=\) \(\ds 12.4\)

$\blacksquare$


Sources

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