The circumcentre of the isoceles triangle ABC is on and `angle ABC=120^(@)`. If the radius of the circle be 5 cm, then determine the length of AB.s

Explanation

Let us join `O,A,O,B and O,C`.
The central angle produced by the arc AC= reflex `angle AOC` and angle in circle `=angle ABC=120^(@)`
By theore-34 reflex `angleAOC=2 angle ABC`.
or, `360^(@)- angleAOC=2 xx 120^(@)`
`or 360^(@)- angleAOC=240^(@) `
or, `angleAOC=360^(@)-240^(@)=120^(@).....(1)`
Again, in `Delta ABC, AB=BC [ :. BC` is isoscele]
Now, in `Delta OAB and Delta OCB, OA=OC [ :.` radii of same circle]
`BA=BC and OB` is common to both.
`:. Delta OAB~=Delta OCB` [ by the condition of S-S-S congruent]
`:. angleAOB=angleBOC [ :.` similar angles of congruent triangles]

Now, `angle ABO+ angleCBO= angleABC=120^(@)`
or, `angleABO+angle ABO=120^(@)[ :. angle OBA= angle OBC]`
`or , 2 angleABO=120^(@), or , angleABO= (120^(@))/(2)= 60^(@)`
Again, `angleAOB+ angle COB= angle AOC= 120^(@)` [ by (1)]
or, `angleAOB+ angleAOB=120^(@) [ :. angleCOB= angle AOB]`
`or, 2 angle AOB=120^(@), or, angle AOB=(120^(@))/(2)= 60^(@)`
`:. "in" Delta OAB,= angle ABO=60^(@)`
`:. Delta AOB` is equilateral.
`:. AB=AO=5 cm [ :. AO`= radius =5 cm]
Hence the required lenght of AB= 5 cm.