Video: Finding the Value of a Trigonometric Function given the Coordinates of the Point of Intersection of the Unit Circle with the Terminal Side of an Angle in Standard Position

The terminal side of βˆ π΄π‘‚π΅ in standard position intersects with the unit circle at the point 𝐡 with coordinates (βˆ’0.8, βˆ’0.6). Find tan βˆ π΄π‘‚π΅.

02:57

Video Transcript

The terminal side of angle 𝐴𝑂𝐡 in standard position intersects with the unit circle at the point 𝐡 with coordinates negative 0.8, negative 0.6. Find tan of angle 𝐴𝑂𝐡.

First of all, to help us understand this problem, I’ve actually done a little sketch of the unit circle. And the unit circle is a circle on our axis here with a radius of one. And we can see it crosses at 𝑦 equals one, π‘₯ equals one, 𝑦 equals negative one, and π‘₯ equals negative one. So our center is at the origin. Okay, we can actually see now that we’ve drawn on angle 𝐴𝑂𝐡. And as we’ve drawn on angle 𝐴𝑂𝐡 here, we’ve got the initial side which is obviously along the π‘₯-axis. And we’ve also got our terminal side. And we know that actually the terminal side of angle 𝐴𝑂𝐡 intersects with the unit circle at point 𝐡, which has the coordinates negative 0.8, negative 0.6. Okay, great. So we’ve got that all now on our diagram.

Okay. So now, how we’re gonna use this diagram? Well, actually, what we’re trying to find in this question is the tan of angle 𝐴𝑂𝐡. And we can actually use this little relationship to help us because the tangent of an angle on the unit circle is equal to the slope. And it’s equal to the slope of that terminal side because it’s the slope that it makes with the origin and actually the unit circle itself. Okay, great. So we can now use that to find the tan of angle 𝐴𝑂𝐡.

And now, to find the slope, we remind ourselves that the slope is equal to the change in 𝑦 over the change in π‘₯. Actually, to help us understand what the change in 𝑦 and change in π‘₯ is gonna be in this problem, I’ve actually drawn this little triangle which represents what we’ve got on the main diagram. And first of all, we can see that the change in 𝑦 is actually going to be negative 0.6 because we’re going from the point zero, zero β€” so our 𝑦-coordinate of zero β€” to the point 𝐡, where our 𝑦-coordinate is negative 0.6. And we could see that our change in π‘₯ is actually going to be negative 0.8. And again, that’s because we’re going from the point zero, zero β€” so our π‘₯-coordinate of zero β€” to a new point 𝐡, where the π‘₯-coordinate is actually negative 0.8.

Okay, great. We’ve got them. And we can substitute them into our equation to find the slope. And just before we actually substitute them back into the equation, just remind ourselves that because we’ve got the slope π‘š and we know from our relationship that that’s gonna be equal to tan of the angle of 𝐴𝑂𝐡, we can therefore say that the tan of angle 𝐴𝑂𝐡 is equal to negative 0.6, so our change in 𝑦, over negative 0.8, our change in π‘₯.

So therefore, if we simplify this, and we can actually simplify this because we can divide the numerator and the denominator by negative 0.2, we can therefore say that the tan of angle 𝐴𝑂𝐡 is equal to three-quarters, three over four.

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