Sine Rule
For any triangle ABC, a = b = c Sin (A) Sin (B) Sin (C)
Where a, b and c are the sides opposite to ∠ A, ∠ B and ∠ C respectively is called sine rule. For proving the result we have two cases:
Case I: The perpendicular height is drawn inside the triangle.
Case II: The perpendicular height is drawn outside the triangle.
Let the sides be AB = c, BC = a and AC = b. Let the perpendicular height be CD = h. In case I the perpendicular CD is inside the triangle and in case II the perpendicular is outside the triangle.
In ∆ACD, Sin (A) = h/b [Case I]
h = b Sin (A) – (i)
Also, in ∆BCD
Sin (B) = h/a [Case I]
Sin (180-B) = h/a [Case II]
Sin (B) = h/a
h = a Sin (B) – (ii)
From (i) and (ii), we have
b Sin(A) = a Sin(B)
b = a [Rearranging the terms]
Sin(B) Sin(A)
Similarly if the perpendicular is drawn from A, we get
b = c
Sin(B) Sin(C)
Combining both the equations we get
a = b = c
Sin(A) Sin(B) Sin(C)
Cosine Rule
For any triangle ABC, a2 = b2 + c2 -2bc Cos (A), where a, b and c are the sides opposite to ∠A, ∠B and ∠C respectively. The equation above can also be written as
Here also there are two cases:
Case I: Perpendicular height is drawn inside the triangle.
Case II: Perpendicular height is drawn outside the triangle.
In ∆ACD, applying Pythagoras’ Theorem, we have
b2 = x2 + h2 – (i)
Cos (A) = x/b
x = b Cos (A) – (ii)
In ∆BCD, Case I, we have
By Pythagoras’ Theorem, a2 = h2+ (c-x)2
In Case II, we have, a2 = h2+ (x-c)2
In both the case on expanding we get,
a2 = h2 + c2 – 2cx + x2
a2 = x2 + h2 + c2 – 2cx
a2 = b2 + c2 – 2c [bcos(A)] [from (i) and (ii)]
a2 = b2 + c2 – 2bc Cos(A)
OR
Thus by drawing the perpendiculars differently we get the remaining formulas such as,
The above mentioned three formulas are known as cosine rule.
Tangent
The law of tangents describes the relationship between the tangent of two angles of a triangle and the lengths of opposite sides. It can be applied to a non-right triangle. The law of tangents is given by,
a - b = tan[(A-B)/2]
a + b tan[(A+B)/2]
Proof: Consider ∆ABC with sides a, b and c which are opposite to the angles A, B and C respectively
To start with the proof, first we must be aware of laws of sines
a = b = d
Sin (A) Sin (B)
a = dsin(A) and b = dsin(B)
a - b = dsin(A) - dsin(B)
a + b dsin(A) + dsin(B)
= d [sin(A) - sin(B)
d [sin(A) + sin(B)]
= sin(A) - sin(B)
sin(A) + sin(B)
= 2 cos[(A+B)/2] [sin(A-B)/2]
2 sin[(A+B)/2] [cos(A-B)/2]
= tan[(A-B)/2]
tan[(A+B)/2]
Hence the law of tangents is proved.
Problems related to these laws
1. In ∆ABC ∠A=106⁰, ∠B=31⁰ and a = 10cm. Solve ∆ABC by calculating ∠C and sides ‘b’ and ‘c’ (round your answers to one decimal place).
Solution: In a triangle ∠ A + ∠ B + ∠ C = 180⁰, ∠ C= 180⁰ - (106⁰ + 31⁰)
= 180⁰ - 137⁰
= 43⁰
We have a/sin(A) = b/sin(B)
10 = b
Sin(106⁰) sin(31⁰)
b = 10 x sin(31⁰)
Sin(106⁰)
= 5.1 cm
Also we have, a = c
sin(A) sin(C)
10 = c
sin(106⁰) sin(43⁰)
c = 10 x sin(43⁰)
Sin(106⁰)
= 7.1cm
2. A triangle has sides equal to 5cm, 10cm and 7cm. Find its angles (Round your answer to one decimal place)
Solution: Let a=10cm, b=7cm and c=5cm
We have a2 = b2 + c2 -2bc Cos (A)
= 49 + 25 – 100
2 x 7 x 5
= -13/35
∠ A= 111.8⁰
= 100 + 25 – 49
2 x 10 x 5
∠ B = 40.5⁰
∠ A + ∠ B + ∠ C = 180⁰
∠ C = 180 – [111.8⁰ + 40.5⁰] = 27.7⁰
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