Law Of Tangents

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 congruence of triangles concept.

Formulas For Laws Of Tangents

Let us assume a right triangle ABC in which side opposite to \(\angle A,\angle B,and \angle C\) are a, b and c respectively. Then, according to the laws of tangent, we have the following three relations :

Laws Of Tangents

\(\frac{a-b}{a+b}=\frac{\tan (\frac{A-B}{2})}{tan(\frac{A+B}{2})}\)

Similarly for other sides,

\(\frac{b-c}{b+c}=\frac{\tan (\frac{B-C}{2})}{tan(\frac{B+C}{2})}\) \(\frac{c-a}{c+a}=\frac{\tan (\frac{C-A}{2})}{tan(\frac{C+A}{2})}\)

Laws Of Tangent Proof

To Prove: \(\frac{a-b}{a+b}=\frac{\tan (\frac{A-B}{2})}{tan(\frac{A+B}{2})}\)

Proof:From the law of Sine,

\(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

Use first and second relation,

\(\frac{a}{\sin A}=\frac{b}{\sin B}=k\) , (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

\(\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A+\sin B}\) ………(1)

Identity Formulas for Sine are:

\(\sin A-\sin B={2\cos \frac{ A+B}{2}}{\sin \frac{A-B}{2}}\) \(\sin A+\sin B={2\sin \frac{ A+B}{2}}{\cos \frac{A-B}{2}}\)

Substitute those formulas in equation (1),we get

\(\frac{a-b}{a+b}=\frac{{2\cos \frac{ A+B}{2}}{\sin \frac{A-B}{2}} }{{2\sin \frac{ A+B}{2}}{\cos \frac{A-B}{2}}}=\frac{\tan \frac{A-B}{2}}{\tan \frac{A+B}{2}}\)

Hence Proved.

Question :

Solve the triangle \(\bigtriangleup ABC\) given a=5,b=3 and \(\angle C\) =\(96^{\circ}\) and find the value of A – B

Solution :

We know that,

\(\angle A+\angle B+\angle C=180^{\circ}\) \(\angle A+\angle B=180^{\circ}-\angle C=180^{\circ}-96^{\circ}=84^{\circ}\)

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,

\(\frac{a-b}{a+b}=\frac{\tan (\frac{A-B}{2})}{tan(\frac{A+B}{2})}\)


\(\Rightarrow \frac{5-3}{5+3}=\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(84^{\circ})}\) \(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{2}{8}\tan 42^{\circ}=0.2251\) \(\Rightarrow \frac{1}{2}(A-B)=12.7^{\circ}\) \(\Rightarrow A-B=25.4^{\circ}\)

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Practise This Question

Match the following table according to the number of unlike terms in an algebraic expression to the appropriate prefix..


a.  3 terms1.Monob.  1 term2.Bic.  8 terms3.Trid.  2 terms4.Poly