# Sine Rule Tutorial

In this tutorial, we are going to learn about sine rule. The circle passing through the three vertices A , B , C of  `Delta`ABC is called the circumcircle . The centre and radius of this circle are called the circum-centre and circum-radius respectively . We know that the perpendicular bisectors of the sides of a triangle are concurrent and the point of concurrence is the circum-centre . We denote this by S and circum radius by R .

Rule :  In  `Delta` ABC  :   `a/sinA`    =    `b/sinB`   =    `c/sinC`    =    2R    ,    where  R  is the circum radius .

Case(1) :  Angle A is acute . See fig(1)

S is the centre of the circumcircle and CD is its diameter .

Then CS = SD = R and CD = 2R .

Join BD

Angle BAC  =  Angle BDC                  ( since angles in the same tangent )

sinA   =  sin(angle BAC)   =  sin(angle BDC)   =  `(BC)/(CD)`   =  `a/(2R)`

`:.`  `a/sinA`    =   2R

Case(2) :  Angle A is the right angle . See figure (2) .

BC  =  a  =  2R  =  2R.1  =  2R sin90o

`:.`  a = 2R sinA  .  Hence  `a/sinA`   =  2R

Case(3) : Angle A is obtuse . See figure(3) .

Angle DBC is right angle .

In the cyclic quadrilateral BACD , angle BDC  = 180o  -  angle BAC   =  180o - A

In `Delta` BDC  ,  sinA  =  sin(180o - A)   =  sin(angle BDC)   =  `(BC)/(CD)`   =  `a/(2R)`

Hence `a/sinA`  =  2R

In a similar way , we can prove  `b/sinB`  =  2R  ,  `c/sinC`   =  2R

`:.`   `a/sinA`   =  `b/sinB`   =  `c/sinC`   =  2R

Note :

• The above rule is called  the  'sine rule' or 'law of sines' . Also in right angled triangle,
=    circum diameter
• a  =  2R sinA  ,  b  =  2R sinB  ,  c  =  2R sinC
(or)   sinA   =  `1/(2R)`   ,   sinB  =  `b/(2R)`   ,  sinC  =  `c/(2R)`

## Solved Problems tutorial on sine rule

1) In `Delta` ABC   if  a  =  3  ,  b  =  4  and  sinA = `3/4`  , find angle B .

Solution : From sine rule  `a/sinA`   =  `b/sinB`

sinB   =   `(bsinB)/a`

=    `(4(3/4))/3`

=    1

`rArr`    B   =   90o

2) In `Delta` ABC   ,  if  a  =  4  ,  c  =  5  and  sinA  =  `4/5`   ,  find angle B

Solution :   From  sine rule  `a/sinA`     =    `c/sinC`

sinC  =   `(csinA)/a`

=    `(5(4/5))/4`

=   1

`rArr`   C  =  90o

## More Solved Problems tutorial on sine rule

1) If  b  =  4 cms ,  A  =  45o  ,  B  =  30o  ,  find  a .

Solution :  From sine , we have   `a/sinA`   =  `b/sinB`

a    =     `(bsinA)/sinB`

=   `(4sin45^o)/sin30^o`

=    `(4(1/sqrt(2)))/(1/2)`

=      `(4/sqrt(2))` `(2)`

a     =     4`sqrt(2)`

2)  Show that  a2 cotA  +  b2 cotB  +  c2 cotC  =  `(abc)/R`

Solution :  L.H.S  =  a2 cotA  +  b2 cotB  +  c2 cotC

=  4R2 sin2A . `cosA/sinA`   +  4R2 sin2B . `cosB/sinB`   +  4R2 sin2C .  `cosC/sinC`             ( by sine rule )

=    2R2 ( 2 sinA cosA + 2 sinB cosB + 2 sinC cosC )

=     2R2 ( sin2A  +  sin2B  +  sin2C )

=    2R2 ( 4 sinA sinB sinC )                                                 ( from identities )

=      `1/R` ( 2R sinA ) ( 2R sinB ) ( 2R sinC )

=    `1/R` (a) (b) (c)

=    `(abc)/R`

=  R.H.S