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Law of Tangents

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Trigonometry is a field of mathematics that determines the angles and unknown lengths of sides of a triangle using trigonometric ratios. There are six trigonometric ratios: sin, cos, tan, cosec, sec and cot. Each ratio displays a specific value for a specific value of the angle. Angles can be expressed in either radians or in degrees. 

Law of tangents

The law of tangents is a trigonometric law that describes the relationship between the sides and angles of a right triangle. The tangent rule describes the link between the sum and differences of a triangle’s sides and angles. The tangent rule can be applied to any triangle with two sides and one angle or one side and two angles to determine the remaining parts. The law of tangents, like the sine and cosine laws, has a wide range of applications in mathematics. The tangents law for a triangle with angles A, B, and C opposing the sides a, b, and c is as follows:

\frac{a-b}{a+b}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}

In simple words, the tangent rule states that the ratio of the difference to the sum of any two given sides of a right triangle is the same as the ratio of the tangent of half the difference to the tangent of half the sum of those sides.

Proof

Let ABC be a right triangle with sides opposite to ∠A, ∠B, ∠C being of the lengths a, b and c respectively. Then, by using the law of sines, we have:

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

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

⇒ a = ksin A and b = ksinB

Thus, a – b = k (sin A – sin B) and a + b = k (sin A + sin B)

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

Since, sinA – sinB = 2cos\frac{A+B}{2}sin\frac{A-B}{2}     and sinA + sinB = 2sin\frac{A+B}{2}cos\frac{A-B}{2}

Substituting the above values in equation (1), we have:

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

⇒ \frac{a-b}{a+b}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}

Hence proved. 

Sample Problems

Problem 1. In triangle ABC, a = 10, b = 7, and ∠C  = 80°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}.

So, \frac{10-7}{10+7}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{84\degree}{2}]}

⇒ \frac{3}{17}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[84\degree]}

⇒ tan\frac{1}{2}(A-B)=\frac{3}{17}tan42\degree=0.40436

⇒ \frac{1}{2}(A-B)=22.01°

⇒ A – B = 44.02°.

Problem 2. In triangle ABC, a = 5, b = 3, and ∠C  = 96°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}   .

So, \frac{5-3}{5+3}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{84\degree}{2}]}\\⇒ \frac{2}{8}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[84\degree]}\\⇒ tan\frac{1}{2}(A-B)=\frac{2}{8}tan42\degree=0.2251\\⇒ \frac{1}{2}(A-B)=12.7°

⇒ A – B = 25.40°.

Problem 3. In triangle ABC, a = 9, b = 3, and ∠C  = 96°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}   .

So, \frac{9-3}{9+3}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{84\degree}{2}]}\\⇒ \frac{6}{12}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[84\degree]}\\⇒ tan\frac{1}{2}(A-B)=\frac{1}{2}tan42\degree=0.4502\\⇒ \frac{1}{2}(A-B)=24.23°

⇒ A – B = 48.46°.

Problem 4. In triangle ABC, a = 8, b = 4, and ∠C  = 80°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}   .

So, \frac{8-4}{8+4}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{100\degree}{2}]}\\⇒ \frac{4}{12}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[50\degree]}\\⇒ tan\frac{1}{2}(A-B)=\frac{1}{3}tan42\degree=0.3972\\⇒ \frac{1}{2}(A-B)=21.66°

⇒ A – B = 43.32°.

Problem 5. In triangle ABC, a = 6, b = 2, and ∠C  = 80°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}   .

So, \frac{6-2}{6+2}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{100\degree}{2}]}\\⇒ \frac{4}{12}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[50\degree]}\\⇒ tan\frac{1}{2}(A-B)=\frac{1}{3}tan42\degree=0.3972\\⇒ \frac{1}{2}(A-B)=12.66°

⇒ A – B = 25.32°.

Problem 6. In triangle ABC, a = 9, b = 3, and ∠C  = 70°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}   .

So, \frac{9-3}{9+3}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{110\degree}{2}]}

⇒ \frac{6}{12}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[110\degree]}

⇒ A – B = 19.02°.

Problem 7. In triangle ABC, a = 10, b = 5, and ∠C  = 96°. Find the value of A – B.

Solution:

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[\frac{A-B}{2}]}{tan[\frac{A+B}{2}]}.

So, \frac{10-5}{10+5}=\frac{tan[\frac{A-B}{2}]}{tan[\frac{84\degree}{2}]}

⇒ \frac{5}{15}=\frac{tan[\frac{A-B}{2}]}{tan\frac{1}{2}[84\degree]}

⇒ tan\frac{1}{2}(A-B)=\frac{3}{17}tan42\degree=0.40436

⇒ A – B = 69.42°.



Last Updated : 24 Jan, 2024
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