Beginning trig.., what is cot in right triangle?

In right triangle ABC, AB = 6, BC = 12, and #mangleA=90^o#. What is the value of #cotC#?

1 Answer
P dilip_k
Jan 24, 2018

In right triangle ABC, AB = 6, BC = 12, and #/_BAC=90^@#

By Pythagoras theorem

#BC^2=AB^2+AC^2#

#=>12^2=6^2+AC^2#

#=>AC=sqrt(144-36)=sqrt108=6sqrt3#

#cotC="adjacent"/"opposite"=(AC)/(AB)=(6sqrt3)/6=sqrt3#

#=>C=cot^-1(sqrt3)=30^@#

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