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-BARSIC- [3]
3 years ago
13

If tan 0=3/4 find csc0

Mathematics
1 answer:
jasenka [17]3 years ago
8 0

Answer:

tanθ = 3/4  p/b

using Pythagoras theorem

H² = P² + b²

H² = 3²+ 4²

H² = 25

H = 5

cosθ = B/H = 4/5

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Solve the triangle A = 2 B = 9 C =8
VARVARA [1.3K]

Answer:

\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

\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}

Then, given the sides a=2, b=9, and c=8.

\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}

For B:

\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}

3 0
1 year ago
Nigel went to a sandwich shop for lunch. The menu to the above shows the types of cheeses and meats available. If Nigel chooses
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Answer:

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Answer:

para el primer dibujo seria 1/12

para el segundo dibujo seria 1/16

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Step-by-step explanation:

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