Geometry!

Tell me how to solve and which rules of geometry to apply. It's sec 4 maths, thanks!

Use opp angles of cyclic quad, use adjacent angles on straight line, cosine rule maybe, sine rule

find area of sector (1/2r^2theta*0.7/2pi) - area of triangle Area(a*b*sin(C)/2)...

I will leave the details another day. You go try first.

a)
angle abc = pi - 1.38 rad

using sine rule,

AB/sin0.7 = 10/sin(pi-1.38)
AB = 6.56cm

b) angle oab = pi - (0.7) - (pi-1.38) = 0.68 rad
area of shaded area = area of sector - area of unshaded triangle

= 1/2 (10^2) (0.7) - 1/2 (10)(6.56)sin(0.68)
= 35 - 20.6244
=14.4 cm^2

Thanks eagle. but part b first line should be:

angle OAB = pi - angle AOB - angle ABO = pi - 0.7 - 1.38 = 1.06159 rad

a)
angle abc = pi - 1.38 rad

using sine rule,

AB/sin0.7 = 10/sin(pi-1.38)
AB = 6.56cm

b) angle oab = pi - (0.7) - (pi-1.38) = 0.68 rad
area of shaded area = area of sector - area of unshaded triangle

= 1/2 (10^2) (0.7) - 1/2 (10)(6.56)sin(0.68)
= 35 - 20.6244
=14.4 cm^2

eagle, think there's a minor typo in part (a) leh.

using sine rule,

AB/sin0.7 = 10/sin1.38

your answer correct >.<

eagle, think there's a minor typo in part (a) leh.

using sine rule,

AB/sin0.7 = 10/sin1.38

your answer correct >.<

yup TS has already pointed out... He's a good self-learner...

the answer appears correct because sin(1.38) gives the same value as sin(pi - 1.38)