(a) Diagram of the angle in Standard position
(b) Length of the hypotenuse (Hyp)
H y p = ( β 5 ) 2 + ( 5 ) 2 = 25 + 25 = 50 H y p = 5 2 Hyp = \sqrt{(-5)^2 + (5)^2}\\
\qquad = \sqrt{25 + 25}\\
\qquad = \sqrt{50}\\
Hyp = 5\sqrt{2} Hy p = ( β 5 ) 2 + ( 5 ) 2 β = 25 + 25 β = 50 β Hy p = 5 2 β
(c) The primary trigonometric ratios are sin, cos and tan.
Since the terminal arm lies in 2nd quadrant, we have to take positive signs for sin, while others are negative.
opp = 5, adj = -5
(i) sin
sin β‘ Ξ± = o p p h y p = 5 5 2 = 1 2 = 1 1.4142 sin β‘ Ξ± = 0.707 \sin{\alpha} = \dfrac{opp}{hyp}=\dfrac{5}{5\sqrt{2}}=\dfrac{1}{\sqrt{2}}=\dfrac{1}{1.4142}\\
\bold{\sin{\alpha} = 0.707} sin Ξ± = h y p o pp β = 5 2 β 5 β = 2 β 1 β = 1.4142 1 β sin Ξ± = 0.707
(ii) cos
cos β‘ Ξ± = a d j h y p = β 5 5 2 = β 1 2 = β 1 1.4142 cos β‘ Ξ± = β 0.707 \cos{\alpha} = \dfrac{adj}{hyp}=\dfrac{-5}{5\sqrt{2}}=-\dfrac{1}{\sqrt{2}}=-\dfrac{1}{1.4142}\\
\bold{\cos{\alpha} = -0.707} cos Ξ± = h y p a d j β = 5 2 β β 5 β = β 2 β 1 β = β 1.4142 1 β cos Ξ± = β 0.707
(iii) tan
tan β‘ Ξ± = o p p a d j = 5 β 5 = β 1 \tan{\alpha} = \dfrac{opp}{adj}=\dfrac{5}{-5}=\bold{-1} tan Ξ± = a d j o pp β = β 5 5 β = β 1