In a circle of radius `r ,` chords of length `aa n dbc m` subtend angles `thetaa n d3theta` , respectively, at the center. Show that `r=asqrt(a/(3a-b))c m`

Explanation

Applying cosine rule in `Delta OAB`, we get
`cos theta = (2r^(2) - a^(2))/(2r^(2))`

or `a^(2) = 2r^(2) (1 - cos theta) " or " a = 2r sin.(theta)/(2)`
Applying cosine rule in `Delta OBC`, we get
`cos 3 theta = (2r^(2) - b^(2))/(2r^(2))`
`rArr b = 2r sin ((3 theta)/(2)) = 2r [3 sin.(theta)/(2) - 4 sin^(3). (theta)/(2)]`
`=2r [(3a)/(2r) - (4a^(3))/(8r^(3))] = 3a - (a^(3))/(r^(2))`
or `r^(2) = (a^(3))/(3a - b)`
or `r = a sqrt((a)/(3a - b)) cm`