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 .
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
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`
`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`
`rArr` C = 90o
1) If b = 4 cms , A = 45o , B = 30o , find a .
Solution : From sine , we have `a/sinA` = `b/sinB`
a = `(bsinA)/sinB`
= `(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)