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 (B−C2) = (b−cb+c) cot A2
(ii) tan (C−A2) = (c−ac+a) cot B2
(iii) tan (A−B2) = (a−ba+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
⇒ (b−cb+c) = sinB−sinCsinB+sinC, [Applying Dividendo and Componendo]
⇒ (b−cb+c) = 2cos(B+C2)sin(B−C2)2sin(B+C2)cos(B−C2)
⇒ (b−cb+c) = cot (B+C2) tan (B−C2)
⇒ (b−cb+c) = cot (π2 - A2) tan (B−C2), [Since, A + B + C = π ⇒ B+C2 = π2 - A2]
⇒ (b−cb+c) = tan A2 tan (B−C2)
⇒ (b−cb+c) = tanB−C2cotA2
Therefore, tan (B−C2) = (b−cb+c) cot A2. Proved.
Similarly, we can prove that the formulae (ii) tan (C−A2) = (c−ac+a) cot B2 and (iii) tan (A−B2) = (a−ba+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 (B−C2) = (b−cb+c) cot A2
R.H.S. = (b−cb+c) cot A2
= ksinB−ksinCksinB+ksinC cot A2, [Using (1)]
= (sinB−sinCsinB+sinC) cot A2
= 2sin(B−C2)cos(B+c2)2sin(B+C2)cos(B−c2)
= tan (B−C2) cot (B+C2) cot A2
= tan (B−C2) cot (π2 - A2) cot A2, [Since, A + B + C = π ⇒ B+C2 = π2 - A2]
= tan (B−C2) tan A2 cot A2
= tan (B−C2) = 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 (A−B2) = (a−ba+b) cot C2 we get,
tan A−B2 = - 1−√31+√3 cot π62
⇒ tan A−B2 = 1−√31+√3 ∙ cot 15°
⇒ tan A−B2 = - 1−√31+√3 ∙ cot ( 45° - 30°)
⇒ tan A−B2 = - 1−√31+√3 ∙ cot45°cot30°+1cot45°−cot30°
⇒ tan A−B2 = - 1−√31+√3 ∙ 1−√31+√3
⇒ tan A−B2 = -1
⇒ tan A−B2 = tan (-45°)
Therefore, A−B2 = - 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.
11 and 12 Grade Math
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