Posted by dena on Monday, January 17, 2011 at 3:20am.
(We know that in rt triangle
Sin=P/H)
given sinA=_/3/2
SO P=_/3 , H=2
ACB is rt triangle
by pythagoras theorem
H^2=P^2+B^2
B^2=H^2-P^2
B^2=(2)^2-(_/3)^2
B^2=4-3
B^2=1
B=_/1
B=1
(where P=perpendicular,
B=base,H=hypotenuse)
thank you.
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