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December 21, 2014

December 21, 2014

Posted by **dena** on Monday, January 17, 2011 at 3:20am.

where in triangle abc c isthert angle and sin a √3/2 - find csc b-please explain where the 1 comes from

- trig -
**Anonymous**, Monday, January 17, 2011 at 9:29am(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)

- trig -
**deena**, Monday, January 17, 2011 at 12:13pmthank you.

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