A central angle of a circle with radius 10 has a measure of 1.35 radians. The secant line between the end points of the arc of the angle divides the sector swept by the arc into two regions, one a triangle and the other bounded by the secant and the arc. What is the area of the region bounded by the secant and the arc?

This is what I have so far:
A1=1/2*r squared*theta, so A1=1/2*10 squared*1.35 rad=67.5

A2=1/2*a*b*sin theta, so A2=1/2*10*?*sin ?

A=A1-A2, so 67.5-??

*I'm not sure how to find b or how about sin theta in A2.... ?

(My scanner isn't working properly so I drew the picture that correlates to the problem below)

2. Originally Posted by accordry
A central angle of a circle with radius 10 has a measure of 1.35 radians. The secant line between the end points of the arc of the angle divides the sector swept by the arc into two regions, one a triangle and the other bounded by the secant and the arc. What is the area of the region bounded by the secant and the arc?

This is what I have so far:
A1=1/2*r squared*theta, so A1=1/2*10 squared*1.35 rad=67.5

A2=1/2*a*b*sin theta, so A2=1/2*10*?*sin ?

A=A1-A2, so 67.5-??

*I'm not sure how to find b or how about sin theta in A2.... ?
Your A1 is the area of the sector.

Your A2 is the area of the isosceles tiangle, where your a and b are radii of the circle, and theta is the included angle between a and b ....or the given central angle 1.35 radians, so,
A2 = (1/2)(10)(10)sin(1.35 rad) = 48.786 sq.units

Hence, area of circular segment is
A = 67.5 - 48.786 = 18.714 sq.units ----answer.