How do I find the area of JKL?
Sorry.....the 'waiting for moderation...' thing is not working again today. I cannot see your diagram.
LJ = 20cos(60)
JK = 20sin(60)
Area = (1/2)*LJ*JK
note that cos(60) = 1/2 and sin(60) = sqrt(3)/2
Currently I’m using an iPad and this particular image is clear. It’s a right-angled triangle, LJK,
angle L is 60deg, hypotenuse LK is 20.
There are other images in other postings which do seem to suffer from the “waiting for moderation” with the iPad.