Question

Question #57a6c

Answer 1

see below

Explanation:

Since `tan alpha =o/a =20/21` using pythagorean theorem we have

`c^2=20^2+21^2`

`c=sqrt(20^2+21^2)=sqrt841=29` Therefore, `cos alpha = 21/29` and `sin alpha = 20/29`

We also need to half `180^@< alpha < 270^@` hence,

`180/2^@< alpha/2 < 270^@/2`

`90^@ < alpha/2 < 135^@`--> Quadrant II

`sin (alpha/2) = sqrt(1/2(1-cosalpha))=sqrt(1/2(1-21/29)` =`sqrt(1/2((29-21)/29)) =sqrt(1/2(8/29))=sqrt(4/29) = 2/sqrt29 :. = (2sqrt29)/29`

`cos (alpha /2)= -sqrt(1/2(1+cosalpha))=-sqrt(1/2(1+21/29))= - sqrt(1/2((29+21)/29) ) = -sqrt(1/2(50/29) )= -sqrt(25/29) = -5/sqrt29 :. = (-5sqrt29)/29`

`tan (alpha / 2) = sin(alpha/2)/cos(alpha/2) = (2/sqrt29)/(-5/sqrt29) = 2/sqrt29 * sqrt29 / -5`

`= 2/cancel sqrt29 * cancel sqrt29 / -5 :. = -2/5`