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xz_007 [3.2K]
2 years ago
5

Tower A is 20 miles due North of tower B. A dinosaur is spotted from the tower A at °E and from tower Bat N34°E. How far is the

dinosaur from tower B?
Mathematics
1 answer:
bekas [8.4K]2 years ago
4 0

Answer:

Step-by-step explanation:

let dinosaur be at C

apply sine formula

∠A=56°

∠B=34°

∠C=180-(56+34)=180-90=90°

\frac{20}{sin 90} =\frac{BC}{sin 56} \\BC=20 sin 56 \approx 16.58 miles

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2 years ago
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}

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1 year ago
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Hope this helps.

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erma4kov [3.2K]
Answer: area= 4.7

Area=pi*r^2=3.14*1.5=4.71=4.7 (nearest tenth)
6 0
2 years ago
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