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rusak2 [61]
3 years ago
13

Find one positive angle whose sine is negative 0.70.

Mathematics
1 answer:
Svetach [21]3 years ago
3 0

Answer:

  • <u><em>Both 2.34 radians and 5.48 radians (approximately) are positive angles whose sine is - 0.70</em></u>

<u><em></em></u>

Explanation:

The <em>sine</em> is the y-coordinate of the point on the unit circle.

The graph shows that, in the first quadrant, the point with y-coordinate = 0.7 has an angle of 0.8 radians (you read it by moving horizontally to the right from the y-axis at the height of 0.70).

Hence, the sine of 0.8 radians is 0.70 (positive).

The sine is negative in the third and fourth quadrants (the y-coordinate is negative below the y-axis).

Then, 0.8 radians is your reference angle.

  • The positive angle in the third quadrant is π radians - reference angle. This is, π radians - 0.8 radians ≈ 3.14 radians - 0.8 radians = 2.34 radians.

  • The positive angle in the fourth quadrant if 2π radians - 0.8 radians ≈ 6.28 radians - 0.8 radians = 5.48 radians.
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Using slope-intercept form, y = mx + b where m = slope and b = y-intercept:

We know our slope is -6. This can be interpreted as -6/1, which rise-over-run-wise, means that when y changes by 6, x changes inversely by 1.
To find that y-intercept, though, we need to find the value of y when x = 0.
Use our point (-9, -3) to find this...
We want to add 9 to x so that it becomes 0.
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Our y-intercept is at (0, -57), with -57 being the value of b we put in our equation.

\boxed{y=-6x-57}

You could also just use point-slope form:
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And convert to slope-intercept if you want:
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answer
<span>d) x = 10, y = 12</span>
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