We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
444
2
avatar+59 

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+7354 
+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+99237 
+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

21 Online Users

avatar
avatar
avatar