Similarity of triangles: (½S-½s)/s = ½s/(Scos30°-s) (S-s)(S√3/2 -s)=s² S²√3/2 - Ss - Ss√3/2 +s²=s² S²√3/2 = Ss(1+√3/2) S/s= (1+√3/2)/(√3/2) S/s =2/√3+1 = 2,1547 tan x= (½S+½s) / ½s = (S+s)/s = S/s + 1 x = 72,412° ( Solved √ ) Given data S=100mm was just for distraction, is a redundancy, is not necessary !!!
@MathandEngineering
8 күн бұрын
Yes you are right, the Question just came that way
@santiagoarosam430
8 күн бұрын
Tg X=[2a+(2a/√3)]/a =4/(3-√3)=3,1547..---> X=72,4120°.. Gracias y saludos
@MathandEngineering
8 күн бұрын
Thank you friend, this is a perfect one
@abdmoh6480
9 күн бұрын
Thank you so much sir for this useful exercise. I've given it a try and I've found that: Tan(X)=(6+2√3)/3 so X=72°.412 keep going sir because you're doing great.
@MathandEngineering
9 күн бұрын
Thank you so much, I am glad you found it useful, and also for the encouragement
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