Question #57a6c

1 Answer
Jan 26, 2017

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#