Law of Tangents

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 (BC2) = (bcb+c) cot A2

(ii) tan (CA2)  = (cac+a) cot B2

(iii) tan (AB2) = (aba+b) cot C2

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

bsinB = csinC

bc = sinBsinC

 ⇒ (bcb+c) = sinBsinCsinB+sinC, [Applying Dividendo and Componendo]

⇒ (bcb+c) = 2cos(B+C2)sin(BC2)2sin(B+C2)cos(BC2)

⇒ (bcb+c) = cot (B+C2) tan (BC2)

⇒ (bcb+c) = cot (π2 - A2) tan (BC2), [Since, A + B + C = π ⇒ B+C2 = π2 - A2]

⇒ (bcb+c) = tan A2 tan (BC2)

⇒ (bcb+c) =  tanBC2cotA2

Therefore, tan (BC2) = (bcb+c) cot A2.                        Proved.

Similarly, we can prove that the formulae (ii) tan (CA2)  = (cac+a) cot B2 and (iii) tan (AB2) = (aba+b)  cot C2.

 

Alternative Proof law of tangents:

According to the law of sines, in any triangle ABC,                     

      asinA = bsinB = csinC

Let, asinA = bsinB = csinC = k

Therefore,

asinA = k, bsinB = k and csinC = k

a = k sin A, b = k sin B and c = k sin C ……………………………… (1)

Proof of formula (i) tan (BC2) = (bcb+c) cot A2

R.H.S. = (bcb+c) cot A2

= ksinBksinCksinB+ksinC cot A2, [Using (1)]

= (sinBsinCsinB+sinC) cot A2

= 2sin(BC2)cos(B+c2)2sin(B+C2)cos(Bc2)

= tan (BC2) cot (B+C2) cot A2

= tan (BC2) cot (π2 - A2) cot A2, [Since, A + B + C = π ⇒ B+C2 = π2 - A2]

= tan (BC2) tan A2 cot A2

= tan (BC2) = L.H.S.

Similarly, formula (ii) and (iii) can be proved.


Solved problem using the law of tangents:

If in the triangle ABC, C = π6, b = √3 and a = 1 find the other angles and the third side.

Solution: 

Using the formula, tan (AB2) = (aba+b) cot C2 we get,

tan AB2 = - 131+3 cot π62

tan AB2 = 131+3 ∙ cot 15°

tan AB2 = -  131+3 ∙ cot ( 45° - 30°)

tan AB2 = - 131+3cot45°cot30°+1cot45°cot30°

tan AB2 = - 131+3131+3 

tan AB2 = -1

tan AB2 = tan (-45°)

Therefore, AB2 = - 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, asinA = csinC

Therefore, 1sin30° = csin30°

c = 1

Therefore, the other angles of the triangle are 120° or, 2π3; 30° or, π6; and the length of the third side = c = 1 unit.

 Properties of Triangles




11 and 12 Grade Math 

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