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Question
Below fig shows a sector of a circle, centre O. containing an angle ๐°. Prove that Perimeter of shaded region is ๐ (tan ๐ + sec ๐ +`(pitheta)/180`− 1)
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Solution

Given angle subtended at centre of circle = ๐
∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]
OAB is right angle triangle
Cos ๐ =`(adj.side)/(hypotenuse) =r/OB`⇒ ๐๐ต = ๐ sec ๐ … … (๐)
tan ๐ =`(opp.side)/(adju.side)=AB/r`⇒ ๐ด๐ต = ๐ tan ๐ … … . (๐๐)
Perimeter of shaded region = AB + BC + (CA arc)
= ๐ tan ๐ + (๐๐ต − ๐๐ถ) +`theta/360^@`× 2๐๐
= ๐ tan ๐ + ๐ sec ๐ − ๐ +`(pithetar)/180^@`
= ๐ (tan ๐ + sec ๐ +`(pitheta)/180^@`− 1)
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