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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(tan╬╕ - (╧А╬╕)/180^circ)).

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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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рдЕрдзреНрдпрд╛рдп 13: Areas Related to Circles - EXERCISE 13.2 [рдкреГрд╖реНрда резрей.реиреж]

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рдЖрд░.рдбреА. рд╢рд░реНрдорд╛ Mathematics [English] Class 10
рдЕрдзреНрдпрд╛рдп 13 Areas Related to Circles
EXERCISE 13.2 | Q 16. (ii) | рдкреГрд╖реНрда резрей.реиреж
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