Question

sin Q = 1/3, dan Q terletak pada kuadran pertama, tentukan cos Q dan tan Q ?

Answer 1

Sin q = 1/3
sin = depan / miring
samping = √3²-1²
= 2√2
cos q = samping /miring
=2√2 / 3
tan Q = depan / samping
= 1 / 2√2
= 2√2 / 8

Answer 2

sin Q = 1/3 ---> sin Q = y / r ---> y = 1 dan r = 3
x = akar ( r² - y² )
= akar [ (3)² - (1)² ]
= akar [ 9 - 1 ]
= √8
= 2√2

cos Q = x / r 
= 2V2 / 3
= 2/3 V3

tan Q = y / x
= 1 / ( 2√2 ) . √2/√2
= √2 / ( 2 . 2 )
= 1/4 √2