circle and trigonomertic

1.area of rectangle = (distance between the centres of 2 circles) x (radius of circle)
= 10 cm x 5 cm = 50 cm^2

area of 2 quarter of a circle =2 x 1/4(pi x radius^2) = 0.5(3.14 x 5^2) = 39.25cm^2shade area = area of rectangle - 2 quarter of a circle = 50cm^2 – 39.25 cm^2 = 10.75 cm^2area of shaded region = 10.75 cm^2 2. 19°35' = 19 degrees 35 minutes = 19.5833° 3 tan θ = (tan 19°35')/(tan12.5°+0.92) (tan 19.5833°)
3 tan θ = --------------------------
(tan12.5°+0.92)

0.355756245
3 tan θ = --------------------------------
(0.221694662 + 0.92)

0.355756245
3 tan θ = -----------------------
1.141694663

tan θ = 0.103867889 θ = 5.930° 3. tan θ = 1/√3 = oppositeside/adjacent sidehypotenuse^2 = (opposite side)^2 + (adjacent side)^2hypotenuse^2 = (1)^2 + (√3)^2hypotenuse^2 = 4hypotenuse = 2sin θ = opposite side/hypotenuse = ½ sin θ = 0.5

1) [10/2 x 10/2 - (10/2)^2 x πx 1/4) x2
= (5x5 - 5^2 xπx 1/4)x2
= (25-6.25π)x 2
= 50 - 12.5π
= 10.75cm

2) 55.81 這條我不太懂

3) θ= 0.5