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The difference and sum of sides of a right triangle and tangents of half of the difference and sum of corresponding angles are described by the rules of tangent (Law of Tan). It depicts the relationship between the length of the opposing sides and the tangent of two triangle angles. The law of tangents, like the laws of sines and cosines, may be applied to a non-right triangle and is just as effective. From the congruence of triangles notion, it may be utilized to discover the remaining sections of a triangle if two angles and one side or two sides and one angle are supplied. These are side-angle-side(SAS) and angle-side-angle(ASA).
Also read: Isosceles Triangle Theorems
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Key Terms: Triangle, Circle, Squares, rectangle, Rhombus, Parallelogram, Trapezium
Law of Tangents Formula
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Assume a right triangle ABC with sides a, b, and c, respectively, opposite to angles A, B, and C. The rules of a tangent then give us the following three relationships:
Triangle ABC
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\) ……(1)
Similarly, for other sides
\(\frac{b-c}{b+c}=\frac{\tan \left(\frac{B-C}{2}\right)}{\tan \left(\frac{B+C}{2}\right)}\) ……(2)
\(\frac{c-a}{c+a}=\frac{\tan \left(\frac{C-A}{2}\right)}{\tan \left(\frac{C+A}{2}\right)}\) ……(3)
We may rewrite the previous law of tangents equation as tan (-) = -tan for any angle since tan (-) = -tan for any angle.
\(\frac{b-a}{b+a}=\frac{\tan \left(\frac{B-A}{2}\right)}{\tan \left(\frac{B+A}{2}\right)}\) ……(4)
Similarly, for other sides
\(\frac{c-b}{c+b}=\frac{\tan \left(\frac{C-B}{2}\right)}{\tan \left(\frac{C+B}{2}\right)}\) ……(5)
\(\frac{a-c}{a+c}=\frac{\tan \left(\frac{A-C}{2}\right)}{\tan \left(\frac{A+C}{2}\right)}\) ……(6)
When a>b, b>c, and c>a the formulas (1), (2), and (3) are used, and when b>a, c>b, and a>c, the formulas (4), (5), and (6) are used.
Read more: Angle Formula
Law of Tangents Proof
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In any triangle ABC,
Read more: Line Segment
Alternative Proof Law of Tangents
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According to the law of sines, in any triangle ABC,
Similarly, formula 2 and 3 can be proved.
Read more: Vertex
Law of Tangents Statements and Reason
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The statements and the reasons related to it are mentioned in table below.
| Statements | Reason |
|---|---|
| \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) | Applying sine rule to the triangle ABC. |
| \(\frac{a}{\sin A}=\frac{b}{\sin B}=d\) | Equating a ratio to a constant. |
| \(\frac{a}{\sin A}=d\) and \(\frac{b}{\sin B}=d\) | Equating each ratio to a constant ‘d’ |
| a = d Sin A and b = d Sin B | Cross Multiplication |
| a + b = d Sin A + d Sin B = d (Sin A + Sin B) → (1) | Sum of ‘a’ and ‘b’ |
| a - b = d Sin A - d Sin B = d (Sin A - Sin B) → (2) | Difference between ‘a’ and ‘b’ |
| \(\sin \mathrm{M}+\sin \mathrm{N}=2 \sin\left(\frac{M+N}{2}\right) \cos \left(\frac{M-N}{2}\right)\) \(\sin \mathrm{M}-\sin \mathrm{N}=2 \cos\left(\frac{M+N}{2}\right) \sin \left(\frac{M-N}{2}\right)\) | Trigonometric Identities |
| \(\frac{a-b}{a+b}=\frac{d(\sin A-\sin B)}{d(\sin A+\sin B)}= \frac{\sin A-\sin B}{\sin A+ \sin B}\) | Dividing 1 and 2 |
| \(\frac{2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)}{2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)}\) | Substituting the trigonometric identities in the above equation |
| \(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\) | By definition, Tan R = Sin R / Cos R |
Also Read: Introduction to Constructions
Things To Remember
- The ratio of the tangent of half the sum and tangent of half the difference of the angles opposite the respective sides is equal to the ratio of the sum and difference of any two sides of a triangle.
- Nasir al-Din al-Tusi, a Persian mathematician, established the law of tangents for triangles in the 13th century. For spherical triangles, he explained the law of tangents.
- The tangent of the difference between two sides and the tangent of their total is equal to the tangent of the half of the difference between their opposing angles and the tangent of half of their sum, according to the spherical law of tangents. The spherical law of tangents is given as follows for a triangle with angles A, B, and C opposing the sides 'a', 'b', and 'c', respectively:
\(\frac{\tan \left(\frac{a-b}{2}\right)}{\tan \left(\frac{a+b}{2}\right)}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
Also read: Difference between Sequence and Series
Sample Questions
Ques. Solve the triangle ABC given a = 5, b = 3, and ∠C = 96° and find the value of A – B. (4 Marks)
Ans. ∠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,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
Therefore,
\(\frac{5-3}{5+3}=\frac{\tan \frac{1}{2} (A-B)}{\tan \frac{1}{2} (84^{\circ})}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{2}{8} \tan 42^{\circ}=0.2251\)
\(\Rightarrow \frac{1}{2}(A-B)=12.7^{\circ}\)
\(\Rightarrow A-B=25.4^{\circ}\)
Ques. If in the triangle ABC, C = π/6, b = √3 and a = 1 find the other angles and the third side. (4 Marks)
Ans.
⇒ B – A = 90° ……………..(1)
Again, A + B + C = 180°
Therefore, A + B = 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, \(\frac{1}{\sin 30^{\circ}}=\frac{c}{\sin 30^{\circ}}\)
c = 1
Therefore, the other angles of the triangle are 120° or, 2π/3; 30° or, π/6; and the length of the third side = c = 1 unit.
Ques. In triangle ABC, a = 10, b = 7, and ∠C = 96°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 80°.
⇒∠A + ∠B= 180°- ∠C = 180° – 96° = 84°
As per the law of tangents
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
So,
\(\frac{10-7}{10+7}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{84}{2}\right)}\)
\(\Rightarrow \frac{3}{17}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \frac{1}{2} (84)}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{3}{17} \tan 42^{\circ}=0.40436\)
\(\Rightarrow \frac{1}{2}(A-B)=22.01^{\circ}\)
\(\Rightarrow A-B=44.02^{\circ}\)
Ques. In triangle ABC, a = 9, b = 3, and ∠C = 96°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 80°.
⇒∠A + ∠B= 180°- ∠C = 180° – 96° = 84°
As per the law of tangents,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
\(\Rightarrow \frac{9-3}{9+3}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left( \frac{84}{2} \right) }\)
\(\Rightarrow \frac{6}{12}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \frac{1}{2} (84)}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{1}{2} \tan 42^{\circ}=0.4502\)
\(\Rightarrow \frac{1}{2}(A-B)=24.23^{\circ}\)
\(\Rightarrow A – B = 48.46^{\circ}\)
Ques. In triangle ABC, a = 8, b = 4, and ∠C = 80°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 80°.
⇒∠A + ∠B= 180°- ∠C = 180° – 80° = 100°
As per the law of tangents,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
\(\Rightarrow \frac{8-4}{8+4}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left( \frac{100}{2} \right) }\)
\(\Rightarrow \frac{4}{12}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \frac{1}{2} (100)}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{1}{3} \tan 50^{\circ}=0.3972\)
\(\Rightarrow \frac{1}{2}(A-B)=21.66^{\circ}\)
\(\Rightarrow A – B = 43.32^{\circ}\)
Ques. In triangle ABC, a = 6, b = 2, and ∠C = 80°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 80°.
⇒∠A + ∠B= 180°- ∠C = 180° – 80° = 100°
As per the law of tangents,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
\(\Rightarrow \frac{6-2}{6+2}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left( \frac{100}{2} \right) }\)
\(\Rightarrow \frac{4}{8}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \frac{1}{2} (100)}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{1}{2} \tan 50^{\circ}=0.5958\)
\(\Rightarrow \frac{1}{2}(A-B)=30.79^{\circ}\)
\(\Rightarrow A – B = 61.58^{\circ}\)
Ques. In triangle ABC, a = 9, b = 3, and ∠C = 70°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 70°.
⇒∠A + ∠B= 180°- ∠C = 180° – 70° = 110°
As per the law of tangents,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
\(\Rightarrow \frac{9-3}{9+3}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left( \frac{110}{2} \right) }\)
\(\Rightarrow \frac{6}{12}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \frac{1}{2} (110)}\)
A – B = 19.02°
Ques. In triangle ABC, a = 10, b = 5, and ∠C = 96°. Find the value of A – B. (4 Marks)
Ans. The angle sum property of a triangle states that ∠A + ∠B + ∠C = 180°.
Since it is given that ∠C = 80°.
⇒∠A + ∠B= 180°- ∠C = 180° – 96° = 84°
As per the law of tangents,
\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)
\(\frac{10-5}{10+5}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{84}{2}\right)}\)
\(\Rightarrow \tan \frac{1}{2}(A-B)=\frac{1}{3} \tan 42^{\circ}=0.30\)
\(\Rightarrow \frac{1}{2}(A-B)=16.7^{\circ}\)
\(\Rightarrow A – B = 33.4^{\circ}\)
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