We will discuss here about the law of tangents or the tangent rule which is required for solving the problems on triangle.
In any triangle ABC,
(i) tan (\(\frac{B - C}{2}\)) = (\(\frac{b - c}{b + c}\)) cot \(\frac{A}{2}\)
(ii) tan (\(\frac{C - A}{2}\)) = (\(\frac{c - a}{c + a}\)) cot \(\frac{B}{2}\)
(iii) tan (\(\frac{A - B}{2}\)) = (\(\frac{a - b}{a + b}\)) cot \(\frac{C}{2}\)
The law of tangents or the tangent rule is also known as Napier’s analogy.
Proof of tangent rule or the law of tangents:
In any triangle ABC we
have
⇒ \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)
⇒ \(\frac{b}{c}\) = \(\frac{sin B}{sin C}\)
⇒ (\(\frac{b - c}{b + c}\)) = \(\frac{sin B - sin C}{sin B + sin C}\), [Applying Dividendo and Componendo]
⇒ (\(\frac{b - c}{b + c}\)) = \(\frac{2 cos (\frac{B + C}{2}) sin (\frac{B - C}{2})}{2 sin (\frac{B + C}{2}) cos (\frac{B - C}{2})}\)
⇒ (\(\frac{b - c}{b + c}\)) = cot (\(\frac{B + C}{2}\)) tan (\(\frac{B - C}{2}\))
⇒ (\(\frac{b - c}{b + c}\)) = cot (\(\frac{π}{2}\) - \(\frac{A}{2}\)) tan (\(\frac{B - C}{2}\)), [Since, A + B + C = π ⇒ \(\frac{B + C}{2}\) = \(\frac{π}{2}\) - \(\frac{A}{2}\)]
⇒ (\(\frac{b - c}{b + c}\)) = tan \(\frac{A}{2}\) tan (\(\frac{B - C}{2}\))
⇒ (\(\frac{b - c}{b + c}\)) = \(\frac{tan \frac{B - C}{2}}{cot \frac{A}{2}}\)
Therefore, tan (\(\frac{B - C}{2}\)) = (\(\frac{b - c}{b + c}\)) cot \(\frac{A}{2}\). Proved.
Similarly, we can prove that the formulae (ii) tan (\(\frac{C - A}{2}\)) = (\(\frac{c - a}{c + a}\)) cot \(\frac{B}{2}\) and (iii) tan (\(\frac{A - B}{2}\)) = (\(\frac{a - b}{a + b}\)) cot \(\frac{C}{2}\).
Alternative Proof law of tangents:
According to the law of sines, in any triangle ABC,
\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\)
Let, \(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\) = k
Therefore,
\(\frac{a}{sin A}\) = k, \(\frac{b}{sin B}\) = k and \(\frac{c}{sin C}\) = k
⇒ a = k sin A, b = k sin B and c = k sin C ……………………………… (1)
Proof of formula (i) tan (\(\frac{B - C}{2}\)) = (\(\frac{b - c}{b + c}\)) cot \(\frac{A}{2}\)
R.H.S. = (\(\frac{b - c}{b + c}\)) cot \(\frac{A}{2}\)
= \(\frac{k sin B - k sin C}{k sin B + k sin C }\) cot \(\frac{A}{2}\), [Using (1)]
= (\(\frac{sin B - sin C}{sin B + sin C }\)) cot \(\frac{A}{2}\)
= \(\frac{2 sin (\frac{B - C}{2}) cos (\frac{B + c}{2})}{2 sin (\frac{B + C}{2}) cos (\frac{B - c}{2})}\)
= tan (\(\frac{B - C}{2}\)) cot (\(\frac{B + C}{2}\)) cot \(\frac{A}{2}\)
= tan (\(\frac{B - C}{2}\)) cot (\(\frac{π}{2}\) - \(\frac{A}{2}\)) cot \(\frac{A}{2}\), [Since, A + B + C = π ⇒ \(\frac{B + C}{2}\) = \(\frac{π}{2}\) - \(\frac{A}{2}\)]
= tan (\(\frac{B - C}{2}\)) tan \(\frac{A}{2}\) cot \(\frac{A}{2}\)
= tan (\(\frac{B - C}{2}\)) = L.H.S.
Similarly, formula (ii) and (iii) can be proved.
Solved problem using the law of tangents:
If in the triangle ABC, C = \(\frac{π}{6}\), b = √3 and a = 1 find the other angles and the third side.
Solution:
Using the formula, tan (\(\frac{A - B}{2}\)) = (\(\frac{a - b}{a + b}\)) cot \(\frac{C}{2}\) we get,
tan \(\frac{A - B}{2}\) = - \(\frac{1 - √3}{1 + √3}\) cot \(\frac{\frac{π}{6}}{2}\)
⇒ tan \(\frac{A - B}{2}\) = \(\frac{1 - √3}{1 + √3}\) ∙ cot 15°
⇒ tan \(\frac{A - B}{2}\) = - \(\frac{1 - √3}{1 + √3}\) ∙ cot ( 45° - 30°)
⇒ tan \(\frac{A - B}{2}\) = - \(\frac{1 - √3}{1 + √3}\) ∙ \(\frac{cot 45° cot 30° + 1}{cot 45° - cot 30°}\)
⇒ tan \(\frac{A - B}{2}\) = - \(\frac{1 - √3}{1 + √3}\) ∙ \(\frac{1 - √3}{1 + √3}\)
⇒ tan \(\frac{A - B}{2}\) = -1
⇒ tan \(\frac{A - B}{2}\) = tan (-45°)
Therefore, \(\frac{A - B}{2}\) = - 45°
⇒ B - A = 90° ……………..(1)
Again, A + B + C = 180°
Therefore, A + 8 = 180° - 30° = 150° ………………(2)
Now, adding (1) and (2) we get, 2B = 240°
⇒ B = 120°
Therefore, A = 150° - 120° = 30°
Again, \(\frac{a}{sin A}\) = \(\frac{c}{sin C}\)
Therefore, \(\frac{1}{sin 30°}\) = \(\frac{c}{sin 30°}\)
⇒ c = 1
Therefore, the other angles of the triangle are 120° or, \(\frac{2π}{3}\); 30° or, \(\frac{π}{6}\); and the length of the third side = c = 1 unit.
11 and 12 Grade Math
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