Find the area of triangle JKL when LK is 20 units, angle L is 60 degrees, angle J 90 degrees
sin θ = opposite / hypotenuse
sin 60° = JK / 20
20 sin 60° = JK
JK = 10√3
cos θ = adjacent / hypotenuse
cos 60° = JL / 20
20 cos 60° = JL
JL = 10
area of JKL = (1/2) * JL * JK = (1/2) * 10 * 10√3 = 50√3 sq units
cos (60) = JL / KL
cos (60) = JL / 20 ⇒ JL = 20*cos (60)
And the area can be found as
(1/2) JL * KL * sin (60) =
(1/2) 20* cos (60) * 20 * sin (60) =
(1/2) 20 * (1/2) * 20 * sin (60) =
(1/4)*400 * √3 / 2 =
(1/8)*400*√3 =
50√3 units^2
+63 
