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LAWS OF SINE, COSINE AND TANGENT

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Sine Rule

For any triangle ABC,     a        b      c  

                             Sin (A)     Sin (B)   Sin (C)

law_sine_1

Where a, b a
nd c are the sides opposite to ∠ A, ∠ B and ∠ C respectively is called sine rule.  For proving the result we have two cases:law_sine_2



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.
law_sine_3
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

                               law_sine_7

Here also there are two cases:


Case I: Perpendicular height is drawn inside the triangle.

               law_sine_4

Case II: Perpendicular height is drawn outside the triangle.

                        law_sine_5

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


                                law_sine_7


Thus by drawing the perpendiculars differently we get the remaining formulas such as,        


                               law_sine_8


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

                                            
                                          law_sine_6


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)


           law_sine_7


                         = 49 + 25 – 100  
                               2 x 7 x 5

                         = -13/35


                    ∠ A= 111.8⁰

 
           law_sine_9


                           =  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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