Figure shows a sector of a circle, centre O, containing an angle `theta`. Prove that (i) Perimeter of the shaded region is `r ( tan theta + sec theta + (pi theta)/180 -1)` (ii) Area of shaded region is `r^2/2 ( tan theta - pi theta/180)`

Fig. 15.17, shows a sector of a circle, centre O , containing an angle thetao , prove that: (FIGURE) (i) Perimeter of the shaded region is r\ (tantheta+sectheta+(pitheta)/(180)-1) (ii) Area of the shaded region is (r^2)/2\ (tantheta-(pitheta)/(180))

Fig. 15.17, shows a sector of a circle, centre O , containing an angle thetao , prove that: (FIGURE) (i) Perimeter of the shaded region is r\ (tantheta+sectheta+(pitheta)/(180)-1) (ii) Area of the shaded region is (r^2)/2\ (tantheta-(pitheta)/(180))

Figure shows a sector of a circle with centre O and angle 'theta' . Prove that: (a) Perimeter of shaded region is r(tan theta+sec theta+ (pi theta)/(180^@)-1) units (b) area of the shaded region is r^2/2 (tan theta-(pi theta)/(180^@))sq. units

Prove: sec theta + tan theta = 1/(sec theta - tan theta)

Figure shows a sector of a circle of radius r containing an angle theta^@ The area of the sector is A cm^2 and the perimeter is 50cm. Prove that theta = (360)/pi ((25)/r-1) and A=25r -r^2

Prove that (1)/(sec theta - tan theta) = sec theta + tan theta

In the given figure, is shown a sector OAP of a circle with centre 0, containing /_ 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 + pi theta/180 - 1]

In the given figure,is shown a sector OAP of a circle with centre 0, containing /_ 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+pi(theta)/(180)-1]

prove that (1)/(sec theta -tan theta) = sec theta + tan theta

Figure shows a sector of a circle of radius r containing an angle theta^(@) The area of the sector is Acm^(2) and the perimeter is 50cm. Prove that theta=(360)/(pi)((25)/(r)-1) and A=25r-r^(2)