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valina [46]
3 years ago
10

Solve the equation sin theta = .3 to the nearest tenth. Use the restrictions 90 < theta <180

Mathematics
1 answer:
olga55 [171]3 years ago
5 0
Sinα=3/10

α=arcsin(3/10)°

α≈17.5°

But since we are restricted to 90°<α<180° the angle must lie in the second quadrant so:

α≈(180-17.5)°

α≈162.5°
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\displaystyle\bf\\\boxed{\bf sin\,y=x}\\\\2x^2+x-1=0\\\\x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1-4\cdot2\cdot(-1)}}{2\cdot2}=\\\\=\frac{-1\pm\sqrt{1+8}}{4}=\frac{-1\pm\sqrt{9}}{4}=\frac{-1\pm3}{4}\\\\x_1=\frac{-1+3}{4}=\frac{2}{4}=\boxed{\bf\frac{1}{2}}\\\\x_2=\frac{-1-3}{4}=\frac{-4}{4}=\boxed{\bf-1}\\\\sin\,y=\frac{1}{2}\\\\\boxed{\bf y_1=\frac{\pi}{6}~~or~~(30^o)}\\\\\boxed{\bf y_2=\frac{5\pi}{6}~~or~~(150^o)}\\\\sin\,y=-1\\\\\boxed{\bf y_3=\frac{3\pi}{2}~~or~~(270^o)}

 

 

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3 years ago
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I hope it will help you!!

6 0
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