Law Of Tangents I Definition, Proof, Formula and Sample Examples

The laws of tangent (Law of Tan) describes the relation between difference and sum of sides of a right triangle and tangents of half of the difference and sum of corresponding angles. It represents the relationship between the tangent of two angles of a triangle and the length of the opposite sides. The law of tangents is also applied to a non-right triangle and it is equally as powerful like the law of sines and the law of cosines. It can be used to find the remaining parts of a triangle if two angles and one side or two sides and one angle are given which are referred to as side-angle-side(SAS) and angle-side-angle(ASA), from the congruence of triangles concept.

To understand the law of tangents in a better way, you need some pieces of information for a general triangle even it may or may not be a triangle. The four cases involved are:

Two sides and one opposite angle
One side and two angles
Three Sides
Two sides and the angle between them

Formulas For Laws Of Tangents

Let us assume a right triangle ABC in which sides opposite to

\begin{array}{l} ∠ A, ∠ B, a n d ∠ C \end{array}

are a, b and c respectively. Then, according to the laws of tangent, we have the following three relations :

Laws of tangent

\begin{array}{l} \frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})} \end{array}

….(1)

Similarly for other sides,

\begin{array}{l} \frac{b - c}{b + c} = \frac{\tan (\frac{B - C}{2})}{t a n (\frac{B + C}{2})} \end{array}

…..(2)

\begin{array}{l} \frac{c - a}{c + a} = \frac{\tan (\frac{C - A}{2})}{t a n (\frac{C + A}{2})} \end{array}

…..(3)

Since tan (-θ)= -tan θ for any angle θ, we can switch the order of letters in the above law of tangents formulas and can be rewritten as:

\begin{array}{l} \frac{b - a}{b + a} = \frac{\tan (\frac{B - A}{2})}{t a n (\frac{B + A}{2})} \end{array}

….(4)

Similarly for other sides,

\begin{array}{l} \frac{c - b}{c + b} = \frac{\tan (\frac{C - B}{2})}{t a n (\frac{C + B}{2})} \end{array}

…..(5)

\begin{array}{l} \frac{a - c}{a + c} = \frac{\tan (\frac{A - C}{2})}{t a n (\frac{A + C}{2})} \end{array}

…..(6)

The formulas (1), (2), and (3) are used when a>b, b>c, and c>a, and the formulas (4), (5) and (6) are used when b>a, c>b and a>c.

Laws Of Tangent Proof

To Prove:

\begin{array}{l} \frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})} \end{array}

Proof:From the law of Sine,

\begin{array}{l} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \end{array}

Use first and second relation,

\begin{array}{l} \frac{a}{\sin A} = \frac{b}{\sin B} = k \end{array}

, (say)

a = k sin A and b = k sin B

From this,

a – b = k (sin A – sin B)

a + b = k (sin A + sin B)

So, we get

\begin{array}{l} \frac{a - b}{a + b} = \frac{\sin A - \sin B}{\sin A + \sin B} \end{array}

………(1)

Identity Formulas for Sine are:

\begin{array}{l} \sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2} \\ \sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2} \end{array}

Substitute those formulas in equation (1),we get

\begin{array}{l} \frac{a - b}{a + b} = \frac{2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \frac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}} \end{array}

Hence Proved.

Practice problem

Question :

Solve the triangle

\begin{array}{l} △ A B C \end{array}

given a=5,b=3 and ∠C=96° and find the value of A – B.

Solution :

We know that,

∠A + ∠B + ∠C = 180°

∠A + ∠B= 180°- ∠C = 180° – 96° = 84°

By law of tangents,

for a triangle ABC with sides a, b and c respective to the angles A , B and C is given by,

\begin{array}{l} \frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{t a n (\frac{A + B}{2})} \end{array}

Therefore,

\begin{array}{l} \Rightarrow \frac{5 - 3}{5 + 3} = \frac{\tan \frac{1}{2} (A - B)}{\tan \frac{1}{2} (84^{\circ})} \end{array}

\begin{array}{l} \Rightarrow \tan \frac{1}{2} (A - B) = \frac{2}{8} \tan 42^{\circ} = 0.2251 \end{array}

\begin{array}{l} \Rightarrow \frac{1}{2} (A - B) = {12.7}^{\circ} \end{array}

A – B = 25.4°

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Practice problem

Question :

Solution :

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