Pg24, q49. How do I figure this out? thanks

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1 Answer
Jan 24, 2018

#"Area"=1/2ab sinC#

#c^2=a^2+b^2-2abcosC#

#C=arccos((a^2+b^2-c^2)/(2ab))#

#"Area"=1/2ab sin(arccos((a^2+b^2-c^2)/(2ab)))#

#color(white)("Area")=1/2ab sin(u)#

#u=arccos((a^2+b^2-c^2)/(2ab))#

#cosu=(a^2+b^2-c^2)/(2ab)#

#sinu=sqrt(1-cos^2u)=sqrt(1-((a^2+b^2-c^2)/(2ab))^2)=sqrt(1-(a^2+b^2-c^2)^2/(2ab)^2)=sqrt(((2ab)^2-(a^2+b^2-c^2))/(2ab)^2)#

#"Area"=1/2ab sqrt(((2ab)^2-(a^2+b^2-c^2)^2)/(2ab)^2)#

#"Area"^2=1/4a^2b^2((2ab)^2-(a^2+b^2-c^2)^2)/(2ab)^2#

#"Area"^2=((2ab)^2-(a^2+b^2-c^2))/16#

#"Area"=sqrt(((2ab)^2-(a^2+b^2-c^2))/16)#

#a=63m#
#b=22m#
#c=55m#

#"Area"=sqrt(((2*63*22)^2-(63^2+22^2-55^2))/16)~~594m^2#

#s=(a+b+c)/2=(63+22+55)/2=70#

#"Area"=sqrt(70(70-63)(70-22)(70-55))~~594m^2#