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Figure, shows a sector of a circle, centre O, containing an angle ЁЭЬГ°. Prove that area of the shaded region is `r^2/2(tanθ - (πθ)/180^circ)`.

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Given angle subtended at centre of circle = ЁЭЬГ
∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]
OAB is right angle triangle
Cos ЁЭЬГ =`(adj.side)/(hypotenuse) =r/OB`⇒ ЁЭСВЁЭР╡ = ЁЭСЯ sec ЁЭЬГ … … (ЁЭСЦ)
tan ЁЭЬГ =`(opp.side)/(adju.side)=AB/r`⇒ ЁЭР┤ЁЭР╡ = ЁЭСЯ tan ЁЭЬГ … … . (ЁЭСЦЁЭСЦ)
Area of shaded region = (area of triangle) – (area of sector)
`= (1/2× OA × AB) −theta/360^@× pir^2`
`=1/2× r × r tan theta −r^2/2[theta/180^@× pi]`
=`r^2/2[tantheta −(pitheta)/180^@]`
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рдкрд╛рда 13: Areas Related to Circles - EXERCISE 13.2 [рдкреГрд╖реНрда резрей.реиреж]
