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Below Fig Shows a Sector of a Circle, Centre O. Containing an Angle ๐œƒยฐ. Prove That Perimeter of Shaded Region is ๐‘Ÿ (Tan ๐œƒ + Sec ๐œƒ +`(Pitheta)/180`โˆ’ 1) - Mathematics

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Question

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

Solution

Given angle subtended at centre of circle = ๐œƒ

∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]

OAB is right angle triangle

Cos ๐œƒ =`(adj.side)/(hypotenuse) =r/OB`⇒ ๐‘‚๐ต = ๐‘Ÿ sec ๐œƒ … … (๐‘–)

tan ๐œƒ =`(opp.side)/(adju.side)=AB/r`⇒ ๐ด๐ต = ๐‘Ÿ tan ๐œƒ … … . (๐‘–๐‘–)

Perimeter of shaded region = AB + BC + (CA arc)

= ๐‘Ÿ tan ๐œƒ + (๐‘‚๐ต − ๐‘‚๐ถ) +`theta/360^@`× 2๐œ‹๐‘Ÿ

= ๐‘Ÿ tan ๐œƒ + ๐‘Ÿ sec ๐œƒ − ๐‘Ÿ +`(pithetar)/180^@`

= ๐‘Ÿ (tan ๐œƒ + sec ๐œƒ +`(pitheta)/180^@`− 1)

 

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Chapter 13: Areas Related to Circles - Exercise 13.2 [Page 25]

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RD Sharma Mathematics [English] Class 10
Chapter 13 Areas Related to Circles
Exercise 13.2 | Q 26.1 | Page 25
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