Without using a calculator, determine all values of A in the interval [ 0 , π 2 ) \left[0,\frac{\pi}{2}\right) [ 0 , 2 π ) with the following trigonometric function value. cos A = 3 2 \cos A=\frac{\sqrt3}{2} cos A = 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z">

Let's give this problem a try. So this problem says without using a calculator, determine all the values of a in the interval from 0 to pi over 2 with the following trigonometric function value. And we're given the cosine of a is equal to the square root of 3 over 2. Now, this problem is actually quite tricky because we can't use a calculator, so we're going to have to think outside the box to solve this. But let's see if using the skills that we've learned from trigonometry so far, we can go about figuring out what this angle a is. So let's try this. Now what I'm first going to do is draw a right triangle so we can visualize this better. And I'm going to go ahead and say that our angle a is right about there. Now to figure out how this relates to the sides of the triangle, we're going to use SOHCAHTOA. SOHCAHTOA is this memory tool that we use to remember how the trigonometric functions relate to the sides of a right triangle. And since I can see that we're dealing with cosine, I'm going to use this middle term here that says that cosine is adjacent over hypotenuse. So if I go ahead and go to our angle, the adjacent side is going to be this, which is square root of 3, and then the hypotenuse is going to be this bottom piece, which is 2. Again, because cosine is adjacent over hypotenuse. Now from here, what I'm going to do is solve for the missing side. And to do this, I can use the pythagorean theorem a squared plus b squared is equal to c squared. Now a and b are always going to be the two legs of the triangle. So I'm gonna go ahead and say that our a is going to be the square root of 3. So we're gonna have the square root of 3 squared, and then b is going to be this side of the triangle, which we do not know. And then this is all going to equal c squared, which is always the hypotenuse. So we'll have the hypotenuse squared, and now we just need to solve for b. Now the square root of 3 squared is just gonna give you 3, and this is plus b squared is equal to 2 squared, which is 4. Now what I can do is subtract 3 on both sides of this equation, and that's going to give us that b squared is equal to 4 minuteus 3 which is 1, and then I can take the square root on both sides of this equation to get that b is equal to just 1. So that means that our missing side b is equal to 1. So now that we found all the sides of the right triangle, there's something that we can observe about the behavior of this triangle. Recall that when dealing with a 30, 60, 90 triangle, the hypotenuse of your triangle is always the short leg multiplied by 2. And recall that also the long leg of a 30, 60, 90 triangle is equal to the short leg multiplied by the square root of 3. Now we can see in this triangle that the short leg is 1. So So what that means is that our hypotenuse would just be 1 times 2 or 2, and that checks out. Now what we also can see is that the long leg is gonna be the short leg multiplied by the square root of 3, and one times the square root of 3 would just be the square root of 3, and that also checks out for the long leg of this triangle. So that means we are dealing with a 30, 60, 90 right triangle. And if this angle would be 60 degrees up here, then that means that our angle a must be 30 degrees. Although we're not asked to give our answer in degrees, because notice the interval is from 0 to pi over 2. Pi means we're dealing with radians. So what we actually want to do is convert this angle to radians by taking this and multiplying it by pi over 180 degrees. We've discussed how to do these conversions in previous videos. Now what that means is that the degrees are going to go away, and we can also cancel one of these zeros. Now 3 is going to be right there, and then 18, that's the same thing as 3 times 6. Because 3 times 6 gives you 18. So we can cancel the 3 with that 3 down there, giving us just pi over 6. So the solution for our angle a is going to be pi over 6 radians, and this is the only solution within the interval from 0 to pi over 2. So that's how you can go about solving this problem. As you can see, it's pretty tricky and there's a lot of steps. But hopefully, you can understand the general intuition as to how we were able to find what this angle is. So hope you found this video helpful. Thanks for watching, and let me know if you have any questions.

8. Trigonometric Functions on Right Triangles

Special Right Triangles

Precalculus

8. Trigonometric Functions on Right Triangles

Special Right Triangles

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Multiple Choice

Without using a calculator, determine all values of A in the interval $\left[0,\frac{\pi}{2}\right)$ with the following trigonometric function value.
$\cos A=\frac{\sqrt3}{2}$

$0$ only

$\frac{\pi}{4}$ only

$\frac{\pi}{6}$ only

$\frac{\pi}{3}$ only

Verified step by step guidance

Understand the problem: We need to find the angle A within the interval [0, π/2) for which the cosine value is √3/2.

Recall the unit circle: The cosine of an angle in the unit circle corresponds to the x-coordinate of the point where the terminal side of the angle intersects the circle.

Identify the reference angle: The cosine value √3/2 is associated with a specific reference angle. Recall that cos(π/6) = √3/2.

Verify the interval: Since we are looking for values in the interval [0, π/2), we need to ensure that the angle π/6 falls within this range.

Conclude the solution: The angle A that satisfies cos A = √3/2 within the interval [0, π/2) is π/6.

4:34m

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Without using a calculator, determine all values of A in the interval [0,π2)\left[0,\frac{\pi}{2}\right)[0,2π) with the following trigonometric function value.cosA=32\cos A=\frac{\sqrt3}{2}cosA=23

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Special Right Triangles practice set

Without using a calculator, determine all values of A in the interval $\left[0,\frac{\pi}{2}\right)$ with the following trigonometric function value.
$\cos A=\frac{\sqrt3}{2}$