Find h to the nearest whole number.

For triangle HJK, j = 31, m∠H = 132 degrees, m∠J = 21 degrees, and m∠K = 27 degrees. Find h to the nearest whole number.

$$\small{\text{ $ \tan{(132)}=\frac{h}{p} \qquad \tan{(27)}=\frac{h}{q} \qquad p+q=31\ m $ }}\\ \small{\text{ $ p=\frac{h}{\tan{(132)}} \qquad q=\frac{h}{\tan{(27)}} \qquad \frac{h}{\tan{(132)}} +\frac{h}{\tan{(27)}}=31\ m $ }}\\ \small{\text{ $ h* \left( \frac{1}{\tan{(132)}} + \frac{1}{\tan{(27)}} \right)=31\ m $ }}\\ \small{\text{ $ h* \left( -0.90040404430 + 1.96261050551 \right)=31\ m $ }}\\ \small{\text{ $ h*1.06220646121 =31\ m $ }}\\ \small{\text{ $ h =\frac{31}{1.06220646121}\ m = 29.1845334520\ m $ }}\\ \small{\text{ h to the nearest whole number: $ h = 29\ m $ }}$$

I think the sine rule might be the simplest approach here:

h/sin(H) = j/sin(J)

$${\mathtt{h}} = {\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{132}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{31}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{21}}^\circ\right)}}} \Rightarrow {\mathtt{h}} = {\mathtt{64.284\: \!458\: \!526\: \!596\: \!557\: \!1}}$$

or h = 64 to the nearest whole number

I've assumed h is opposite angle H, j is opposite angle J etc.

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