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Marat540 [252]
1 year ago
14

Point b has coordinates (3,-4) and lies on the circle whose equation is x^2 + y^2= 25. If angle is drawn in a standard position

with its terminal ray extending through point b, what is the sine of the angle?
Mathematics
1 answer:
lakkis [162]1 year ago
3 0
<span>Point B has coordinates (3,-4) and lies on the circle. Draw the perpendiculars from point B to the x-axis and y-axis. Denote the points of intersection with x-axis A and with y-axis C. Consider the right triangle ABO (O is the origin), by tha conditions data: AB=4 and AO=3, then by Pythagorean theorem:
</span>
<span>BO^2=AO^2+AB^2 \\ BO^2=3^2+4^2  \\ BO^2=9+16  \\ BO^2=25  \\ BO=5.
</span>
{Note, that BO is a radius of circle and it wasn't necessarily to use Pythagorean theorem to find BO}
<span>The sine of the angle BOA is</span>
\sin \angle BOA= \dfrac{AB}{BO} = \dfrac{4}{5} =0.8

Since point B is placed in the IV quadrant, the sine of the angle that is <span> drawn in a standard position with its terminal ray will be </span>
<span /><span>
</span><span>
</span>\sin \theta=-0.8 .





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Notice that even tough our polynomial is written in descending form, x^3 is missing, so we are going to use 0x^3 to fill the gap:

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Step 4. Bring down the next term in the dividend and repeat the process for the remaining terms.

After finish the process (check the attached picture), we can conclude that the quotient of 3x^4-4x^2+8x-1 ÷ (x-2) is 3x^3+6x^2+8x+24 with a remainder of 47, or in a different notation: 3x^3+6x^2+8x+24+\frac{47}{x-2}

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