Law of Tangents/Corollary
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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 $C$.
Then:
- $\tan \dfrac {A - B} 2 = \dfrac {a - b} {a + b} \cot \dfrac C 2$
Proof 1
\(\ds \dfrac {a + b} {a - b}\) | \(=\) | \(\ds \dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} }\) | Law of Tangents | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tan \frac {A - B} 2\) | \(=\) | \(\ds \dfrac {a - b} {a + b} \tan \frac {A + B} 2\) | algebraic manipulation | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a - b} {a + b} \tan \frac {180 \degrees - C} 2\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a - b} {a + b} \map \tan {90 \degrees - \frac C 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {a - b} {a + b} \cot \dfrac C 2\) | Tangent of Complement equals Cotangent |
$\blacksquare$
Proof 2
\(\ds \dfrac {a - b} {a + b}\) | \(=\) | \(\ds \dfrac {2 R \sin A - 2 R \sin B} {2 R \sin A + 2 R \sin B}\) | Law of Sines | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \cos \frac {A + B} 2 \sin \frac {A - B} 2} {2 \sin \frac {A + B} 2 \cos \frac {A - B} 2}\) | Sine minus Sine, Sine plus Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan \frac {A - B} 2} {\tan \frac {A + B} 2}\) | Tangent is Sine divided by Cosine and simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan \frac {A - B} 2} {\tan \frac {180 \degrees - C} 2}\) | Sum of Angles of Triangle equals Two Right Angles | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan \frac {A - B} 2} {\map \tan {90^\circ - \frac C 2} }\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\tan \frac {A - B} 2} {\cot \dfrac C 2}\) | Tangent of Complement equals Cotangent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {a - b} {a + b} \cot \dfrac C 2\) | \(=\) | \(\ds \tan \frac {A - B} 2\) | Tangent of Complement equals Cotangent |
Hence the result.
$\blacksquare$
Also presented as
The Corollary to the Law of Tangents can also be found presented as:
- $\dfrac {a - b} {a + b} = \tan \dfrac {A - B} 2 \tan \dfrac C 2$
Also known as
Some sources present this as the Law of Tangents itself.
Examples
Arbitrary Example $1$
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:
\(\ds A\) | \(=\) | \(\ds 96 \degrees 50'\) | ||||||||||||
\(\ds B\) | \(=\) | \(\ds 41 \degrees 10'\) | ||||||||||||
\(\ds c\) | \(=\) | \(\ds 12.4\) |
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Formulae $(37)$
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: Exercises $\text {XXXII}$: $13$.
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): triangle (iv): The tangent rule
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): triangle (iv): The tangent rule