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If possible, find the area of the triangle defined by the following: a = 7, b = 4, y = 43°9.5 square units19.3 square units14 square units16.8 square units

Sagot :

[tex]\begin{gathered} \frac{\text{ A }}{\sin\text{ A}}\text{ = }\frac{B}{\sin \text{ B}} \\ \frac{70}{\sin\text{ A}}\text{ = }\frac{100}{\sin \text{ 35}} \\ \sin \text{ A = }\frac{70\sin 35}{100} \\ \sin \text{ A = 40.15/ 100} \\ \sin \text{ A = 0.40} \\ \text{ A = 23.7\degree} \end{gathered}[/tex]

So C = 180 - 23.7 - 35

= 121.3°

[tex]\begin{gathered} \text{ }\frac{C\text{ }}{\sin\text{ C}}\text{ = }\frac{B}{\sin \text{ B}} \\ \frac{C}{\sin\text{ 121.3}}\text{ = }\frac{100}{\sin \text{ 35}} \\ \text{ C = }\frac{100\sin \text{ 121.3}}{\sin \text{ 35}} \\ C\text{ = }\frac{85.44}{0.57} \\ C\text{ = 150 mi} \end{gathered}[/tex]

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