Triangle CUB has angles C = 66 and B = 80 and side BC = 5 cm. What is the area of the triangle?

If h is the height, then

 

h/tan(80°) + h/tan(66°) = 5   so  h = 5/(1/tan(80°) + 1/tan(66°))

 

The area is just (1/2)*5*h

 

$${\mathtt{Area}} = {\frac{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{5}}}{\left({\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{80}}^\circ\right)}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{66}}^\circ\right)}}}\right)}} \Rightarrow {\mathtt{Area}} = {\mathtt{20.110\: \!829\: \!461\: \!342\: \!909\: \!5}}$$

 

 

Area ≈ 20.11 cm²

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