# Thread: Finding the value of the angle theta in radians, knowing sec theta and cot theta

1. ## Finding the value of the angle theta in radians, knowing sec theta and cot theta

Hi,

The question I have is:

Given that $\displaystyle \sec \theta = -2$, $\displaystyle \cot \theta = -1/3\sqrt{3}$ and $\displaystyle -\pi < 0 < \pi$, find the exact value of the angle $\displaystyle \theta$ in radians. Justify your answer.

Is the following along the right lines?

$\displaystyle \sec\theta=1/cos\theta$ therefore $\displaystyle \cos\theta=1/\sec \theta=-1/2$
$\displaystyle \cot\theta=1/\tan\theta$ therefore $\displaystyle \tan\theta=1/\cot\theta=-3 \sqrt{3}/3$

$\displaystyle \sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}$

Since both $\displaystyle \tan\theta$ and $\displaystyle \cos\theta$ are negative and lie in the range $\displaystyle -\pi < 0 < \pi$, then $\displaystyle \theta$ must lie in the 2nd quadrant so is positive.

2. Originally Posted by cozza
Hi,

The question I have is:

Given that $\displaystyle \sec \theta = -2$, $\displaystyle \cot \theta = -1/3\sqrt{3}$ and $\displaystyle -\pi < 0 < \pi$, find the exact value of the angle $\displaystyle \theta$ in radians. Justify your answer.

Is the following along the right lines?

$\displaystyle \sec\theta=1/cos\theta$ therefore $\displaystyle \cos\theta=1/\sec \theta=-1/2$
$\displaystyle \cot\theta=1/\tan\theta$ therefore $\displaystyle \tan\theta=1/\cot\theta=-3 \sqrt{3}/3$

$\displaystyle \sin\theta =\tan\theta\cos\theta=-3\sqrt{3}/3x-1/2=1/2sqrt{3}$

Since both $\displaystyle \tan\theta$ and $\displaystyle \cos\theta$ are negative and lie in the range $\displaystyle -\pi < 0 < \pi$, then $\displaystyle \theta$ must lie in the 2nd quadrant so is positive.

Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\displaystyle \pi/3$ radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of $\displaystyle \pi/3$ and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the $\displaystyle \pi/6$ angle is 1/2.

3. Originally Posted by HallsofIvy
Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\displaystyle \pi/3$ radians. If you drop a perpendicular from one vertex to the opposite side, you divide the triangle into two congruent right triangle with angles of $\displaystyle \pi/3$ and [/tex]\pi/6[/tex]. If you take the equilateral triangle to have sides of length 1, the hypotenuse of the right triangles is 1 and the leg opposite the $\displaystyle \pi/6$ angle is 1/2.

Thank you for such a quick reply! So does the angle $\displaystyle \theta=1/2sqrt{3}$? Can I then justify this answer by explaining what you have written? Why do I need to know the lenths of the sides in the right angle traingle to find the exact value of the angle $\displaystyle \theta$ in rtadians? Sorry if this seems like a really stupid question.

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### If tan 45° = Cot θ , the the value of θ, in radians is

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