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3 years ago

Please help ASAP failing class if i dont answer correct

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

8 0

(See the imagine for reference)
Let’s solve where they have a triangle, so the height is 9 cm, the base is 3 cm:

1/2 • 9 • 3 = 13.5

Since there’s 2 triangles we do:

13.5(2) = 27

Now the rectangle in the middle, where the height is 9cm and the base is 12cm:

12 • 9 = 108

Add up the areas:

108 + 27 = 135

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$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

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$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

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