a sector of OAP of a circle with centre O, containing angle theta. AB is perpendicular to the radius OA and meets OP produced at B. prove that the perimeter of shaded region is r[tan theta + sec theta + ¶theta/180 -1]

tan thetha = AB/OA 
AB = OA tan thetha = r tan thetha
cos thetha =  r/OB
OB = r/ cos thetha = r sec thetha
​therefore BP = r sec thetha - r
PA = pie. r. thetha/180
therefore required perimeter = r tan thetha + pie. r. thetha/180 + r sec thetha -r = r( r tan thetha = pie. r. thetha/180 = sec thetha - 1)

proved.
 
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After refraction a part of incident Ray of light emerges parallel to the principal axis
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Easy
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Easy method

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Solutioon

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We have angle AOB is theta =>tan theta =AB/r [OA=r (given) ] also, Sec theta =OB/r OB can be written as OP+PB OB =r+PB => sec theta =r+PB/r rsec theta =r+PB rsec theta -r =PB Perimeter of arc AP =2πr theta /360 Now, Perimeter of the region =AB +BP +arc AP => rtan theta +rsec theta-r +2πtheta /360 =>r[tan theta +sec theta +π theta /180-r]
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Easy

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Solution

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