Question 1193620: Given: right △RST with
RT = 8radical 2
and m∠
STV = 150°
Find: RS and ST
△R S T has vertices R on the left, S on top, and T on the right. ∠S is a right angle. A point labeled V lies to the right of T. The horizontal line segment that connects R and T continues to the right and ends at V.
simplest radical form RS =
approximation RS =
simplest radical form ST =
approximation ST =
Found 2 solutions by Boreal, MathTherapy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! So the right triangle is a 30-60-90 one with angle RTS=30 (supplement of the 150 deg)
If RT is 8 sqrt(2) then ST, the hypotenuse of a 30-60-90 rt triangle, is twice that or 16 sqrt (2)
RS is therefore 8 sqrt(2)*sqrt(3)= 8 sqrt(6)=19.60
ST is 22.63
Answer by MathTherapy(10557)  (Show Source):
You can put this solution on YOUR website!
Given: right △RST with
RT = 8radical 2
and m∠
STV = 150°
Find: RS and ST
△R S T has vertices R on the left, S on top, and T on the right. ∠S is a right angle. A point labeled V lies to the right of T. The horizontal line segment that connects R and T continues to the right and ends at V.
simplest radical form RS =
approximation RS =
simplest radical form ST =
approximation ST =
Based on the description, right △RST has its right angle at S. Also, ∡STR = 30o since it's supplementary to straight ∡RTV.
We therefore have a 30-60-90 special △ with RT as its hypotenuse.
As △RST is a 30-60-90 special △ with hypotenuse RT being 8√2, RS, the 
As △RST is a 30-60-90 special △ with hypotenuse RT being 8√2, ST, the 
That's IT!! Nothing MORE, nothing LESS!
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