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Brilliant_brown [7]
4 years ago
12

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. B = 46°, a = 12, b = 11

Mathematics
1 answer:
andreyandreev [35.5K]4 years ago
6 0
*Hint: The Law of Sines is Sin A/a = Sin B/b = Sin C/c

In order to solve this equation, you will have to use this equation:

A= sin^(-1)[a sinB/b]
A= sin^ (-1) [12 (sin 46°) / 11]
A = sin^ (-1) [8.632077604/11]
A = sin ^(-1) [0.7847343276]
A = 51.69611349

Therefore, Sine A would be about 52°
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Answer:

Step-by-step explanation:

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\bar T (t)=\frac{\bar r'(t)}{11\bar r(t)11} =\frac{(7t\cos t)i+(7t\sin t)j}{7t} \\\\\barT(t)=(\cos t)i+(\sin t)j

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\bar N(t)=\bar T'(t)=\frac{(-\sin t)i+(\cos t)j}{(1)} \\\\ \large \boxed {\bar N(t)=(-\sin t)i+(\cos t)j}

K(t)=\frac{|\b\r T'(t)|}{\bar r (t)|} \\\\=\frac{|-\sin t i+\cos t j|}{|7t\cos t +7t \sin t j|}

Using eq (1) and (2)

K(t)=\frac{\sqrt{(-\sin t)^2+(\cos t)^2} }{\sqrt{(7t\cos t)^2+(7t\sin t)^2} }\\\\=\frac{\sqrt{\sin^2 t+\cos^2t} }{\sqrt{49t^2(\cos^2 t+\sin^2t)} }\\\\=\frac{\sqrt{1} }{\sqrt{49t^2\times 1} }  \\\\ \large \boxed {K(t)=\frac{1}{7t} }

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