another way is join centre O and R..In triangle ROS angleROS is 90-2a.because angle ORS is 90 degrees because RS is tangent and OR is line joining centre and point of contact angle POR is 90+2a Which is the arc PR subtends at centre.Then triangle PRO is an equilateral triangle because two sides are radii.Now anglePRO is 45-a..Call intersecting point of lines QS and RO asT...IN triangleTRS all angles a+90+45-a+2.5a= 180 BY solving this a=18degrees. So PR arc subtend an angle 90+2(18)=126degrees.Then arcPR=126/360 of CIrcle circumference 2.22/7multiply by radius 7.
A= 77 m²= ½πR² --> R= 7,0014m
Right triangle RSO:
2β +2a = 90° = 2(β+a) = 2x
x= 45°= 2,5.a --> a= x/2,5= 18°
2β = 90° - 2a = 54°
Anglular sector POR:
α = 180° - 2β = 126°
Arc = α.R = 126°(π/180°).R
Arc = 15,397 cm ( Solved √ )
another way is join centre O and R..In triangle ROS angleROS is 90-2a.because angle ORS is 90 degrees because RS is tangent and OR is line joining centre and point of contact angle POR is 90+2a Which is the arc PR subtends at centre.Then triangle PRO is an equilateral triangle because two sides are radii.Now anglePRO is 45-a..Call intersecting point of lines QS and RO asT...IN triangleTRS all angles a+90+45-a+2.5a= 180 BY solving this a=18degrees. So PR arc subtend an angle 90+2(18)=126degrees.Then arcPR=126/360 of CIrcle circumference 2.22/7multiply by radius 7.
friend, this is a very perfect method, thanks for sharing