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Find the area of triangle JKL when LK is 20 units, angle L is 60 degrees, angle J 90 degrees

 Dec 16, 2017
 #1
avatar+8866 
+2

 

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

 Dec 17, 2017
 #2
avatar+107156 
+2

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

 

 

cool cool cool

 Dec 18, 2017

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