How do you solve the $△ H J K$ $triangle HJK$ given $H = 53^{\circ}, J = 20^{\circ}, h = 31$ $H=53^circ, J=20^circ, h=31$ ?

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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

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

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