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

**cozza** · March 29th 2010, 05:50 AM

Hi,

The question I have is:

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

Is the following along the right lines?

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

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

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

Thanks for your help

**HallsofIvy** · March 29th 2010, 06:18 AM

Originally Posted by cozza

Hi,

The question I have is:

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

Is the following along the right lines?

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

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

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

Thanks for your help

Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\pi/3$ 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 $\pi/3$ 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 $\pi/6$ angle is 1/2.

**cozza** · March 29th 2010, 06:38 AM

Originally Posted by HallsofIvy

Good. Now you also need to know some basic angles. All three angles in an equilateral triangle have measure $\pi/3$ 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 $\pi/3$ 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 $\pi/6$ angle is 1/2.

Thank you for such a quick reply! So does the angle $\theta=1/2sqrt{3}$ ? 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 $\theta$ in rtadians? Sorry if this seems like a really stupid question.

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