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

Question

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

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Answer 1 · 2018-01-24T04:20:54

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

By Pythagoras theorem

$BC^2=AB^2+AC^2$

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

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

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

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