SOLUTION: 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 p

Algebra ->  Formulas -> SOLUTION: 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 p      Log On


   



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) About Me  (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(10552) About Me  (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!