Question

Question #68ba5

Answer 1

Given `OA=OB=6cm and AB=8cm`

(a) `cos/_AOB=(OA^2+OB^2-AB^2)/(2xxOAxxOB)`

`=>cos/_AOB=(6^2+6^2-8^2)/(2xx6xx6)=1/9`

`=>/_AOB=cos^-1(1/9)=83.6^@~~1.46rad`

(b) The length of the minor

`arc AB=OAxx/_AOB=6xx1.46=8.76` cm

(c) The area of the yellow sector
`=pixxOA^2-pixxOA^2xx83.6^@/360^@` `cm^2`

`=22/7xx36xx0.77=87.12cm^2`

Answer 2

a) Angle `/_(AOB) = 1.4595^c = 83.62^0`

b) Minor `arc AB = 8.757 cm`

c) Area of major sector `A_s = 86.8263 cm^2`

Explanation:

a)
Let `/_(AOB) = theta`

`OA = (AB) / (2 sin(theta/2)`

`sin (theta/2) = (AB) / (2 * (OA)) = 8 / (2*6) = 2/3`

`/_(AOB) = theta = 2 * sin^(-1) (2/3) =color(green)( 1.4595^c)`

b)
Length of minor `arc AB = ((/_(AOB) )/ (2pi)) * 2 pi r`

Minor `arc AB = ((1.4595)/(2pi)) * 2pi * 6 = color(brown)(8.757)` `color(brown)(cm^2)`

c)
Area of yellow sector `A_s = ((2pi - theta)/(2pi)) * pi r^2`

`A_s = (2pi - 1.4595)/ (2 (cancel(pi))) * cancel(pi) * 6^2`

`A_s = 18 *(2pi - 1.4595) = color(blue)(86.8263)` `color(blue)(cm^2)`