How do you solve the #triangle HJK# given #H=53^circ, J=20^circ, h=31#?

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Answer 1 · 2017-03-05T09:51:17

#hatK=107°#
#j~=13.28#
#k~=37.12#

#hatK=180°-53°-20°=107°#

Let's apply the sine theorem:

#h/(sin hatH)=j/(sin hatJ)#

and find

#j=(h*sin hatJ)/sin hatH=(31sin20°)/(sin53°)~=13.28#

Again

#h/(sin hatH)=k/(sin hatK)#

and find

#k=(h*sin hatK)/sin hatH=(31sin107°)/(sin53°)~=37.12#