Question Video: Finding the Value of a Trigonometric Function given the Coordinates of the Point of Intersection of the Terminal Side and the Unit Circle Mathematics

Video Transcript

The terminal side of 𝜃 in standard position intersects with the unit circle at point 𝐵, with coordinates eight seventeenths, fifteen seventeenths. Find sec of 𝜃.

First, let’s sketch out this image. Now that we have this rough outline, we can graph point 𝐵, an eight seventeenths, fifteen seventeenths. The terminal side of our 𝜃 intersects point 𝐵. Because we know that this angle is in standard position, its initial side will be the positive 𝑥-axis. And here is our 𝜃.

We’re interested in the sec of 𝜃. We can find the secant ratio by taking the hypotenuse over the adjacent side length. To do this though, we need a right angle. We can draw a perpendicular line down from point 𝐵 to create a right angle. And because we know how the coordinates work, we know each side length of our triangle. The two smallest sides of this triangle are sides eight seventeenths and fifteen seventeenths.

To solve for secant, we need to know the length of the hypotenuse. The Pythagorean theorem can help us solve that. Eight seventeenths squared plus fifteen seventeenths squared equals 𝑐 squared. When we add eight seventeenths squared plus fifteen seventeenths squared, we get the whole number one. 𝑐 squared is equal to one. And so we take the square root of one. The square root of one is one. And our hypotenuse here is length one.

Now if you were paying close attention, you would know that we’re dealing with the unit circle. And because point 𝐵 is located on the unit circle and our vertex is located at point zero, zero, we already know that the hypotenuse would be equal to one, because it is a radius of the unit circle. Any angle who has a terminal side on the unit circle and an initial side on the vertex has a hypotenuse of one. But if you didn’t remember that fact, you can still use the Pythagorean theorem.

Okay, back to our problem, we want to know what the secant of this 𝜃 is. Our hypotenuse is one, and the measure of the adjacent side length is eight over 17. Sec of 𝜃 equals one over eight over 17. But we need to do some simplification here. One divided by eight over 17 is the same thing as one times 17 over eight. One times 17 over eight equals 17 over eight. And so the secant of our angle equals 17 over eight.