# Question Video: Using the Pythagorean Identities and Unit Circle to Solve an Equation in One Variable Mathematics • 10th Grade

The terminal side of angle π in standard position intersects with the unit circle at the point (2π, 3π) where 0 < π < π/2. Find the value of π.

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### Video Transcript

The terminal side of angle π in standard position intersects with the unit circle at the point two π, three π, where zero is less than π is less than π over two. Find the value of π.

First, letβs identify the key information in this question. Weβre told that angle π is in standard position. What does this mean? Standard position refers to the position of an angle within a rectangular coordinate system. An angle is in standard position if its vertex is at the origin and its initial side lies on the positive part of the π₯-axis. We need to consider where the terminal side of this angle is.

The question tells us that π is between zero and π over two radians, which means the terminal side of this angle must be in the first quadrant. Weβre also given some other information about the terminal side of this angle, which is that it intersects with the unit circle at the point two π, three π.

The unit circle is the circle with center at the origin and a radius of one unit. Its equation is π₯ squared plus π¦ squared is equal to one. As the terminal side of angle π intersects the unit circle at the point with coordinates two π, three π, then this point is on the circle, which means its coordinates satisfy its equation.

Substituting two π for π₯ and three π for π¦ into the equation of the unit circle, we have that two π all squared plus three π all squared is equal to one. This gives an equation we can solve in order to find the value of π. Squaring each of the terms gives four π squared plus nine π squared is equal to one.

You need to watch out for a common mistake here, which is forgetting to square the two and the three and thinking instead that the equation should be two π squared plus three π squared is equal to one. This is an incorrect equation and would lead to an incorrect value of π.

Returning to the correct equation and simplifying the left-hand side, we have that 13π squared is equal to one. Dividing both sides of the equation by 13 gives π squared is equal to one over 13. And next, we need to take the square root of both sides of the equation. This gives π is equal to plus or minus the square root of one over 13.

Now the laws of surds tell us that if we take in the square root of a fraction, then this is equal to the square root of the numerator over the square root of the denominator. The square root of one is just one. So we have π is equal to plus or minus one over the square root of 13.

Now this would suggest that there are two possible values for π: positive one over root 13 and negative one over root 13. But letβs check whether both of these are valid. Remember that π is between zero and π over two, which means that itβs situated in the first quadrant. This means that both of its coordinates are positive values. As two π and three π must, therefore, both be greater than zero, this means we need to take only the positive value of π. The value of π is one over the square root of 13.