The point (-9, 40) is on the terminal arm of an angle, $\theta$θ in standard position.
a. what is the length of the line segment connecting the orgin to the point (-9,40)
b. Determine the measure of $\theta$θ , to the nearest tenth of a degree.
6 years ago
The first part of this problem can be solved by aimple application of the Pythagorean theorem:
$x=\sqrt{9^2+40^2}$x=√92+402
$x=41$x=41
For the second part, you take the inverse tangent of the ratio of the sides:
$\tan\alpha=\frac{40}{9}$tanα=409
$\alpha=\tan^{-1}\left(\frac{40}{9}\right)$α=tan−1(409 )
$\alpha=77.3^{\circ}$α=77.3?
To account for the angle lying in the third quadrant, add 180 degrees.
$\theta=180^{\circ}+\alpha$θ=180?+α
$\theta=257.3^{\circ}$θ=257.3?
6 years ago
Answered By Emily H
The first part of this problem can be solved by aimple application of the Pythagorean theorem:
$x=\sqrt{9^2+40^2}$x=√92+402
$x=41$x=41
For the second part, you take the inverse tangent of the ratio of the sides:
$\tan\alpha=\frac{40}{9}$tanα=409
$\alpha=\tan^{-1}\left(\frac{40}{9}\right)$α=tan−1(409 )
$\alpha=77.3^{\circ}$α=77.3?
To account for the angle lying in the third quadrant, add 180 degrees.
$\theta=180^{\circ}+\alpha$θ=180?+α
$\theta=257.3^{\circ}$θ=257.3?
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