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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Solution of triangles