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

The terminal side of angle 𝜃 in standard position intersects with the unit circle at the point (2𝑎, 3𝑎) where 0 < 𝜃 < 𝜋/2. Find the value of 𝑎.

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### Video Transcript

The terminal side of angle 𝜃 in standard position intersects with the unit circle at the point two 𝑎, three 𝑎, where zero is less than 𝜃 is less than 𝜋 over two. Find the value of 𝑎.

First, let’s identify the key information in this question. We’re told that angle 𝜃 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 𝑥-axis. We need to consider where the terminal side of this angle is.

The question tells us that 𝜃 is between zero and 𝜋 over two radians, which means the terminal side of this angle must be in the first quadrant. We’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 𝑎, three 𝑎.

The unit circle is the circle with center at the origin and a radius of one unit. Its equation is 𝑥 squared plus 𝑦 squared is equal to one. As the terminal side of angle 𝜃 intersects the unit circle at the point with coordinates two 𝑎, three 𝑎, then this point is on the circle, which means its coordinates satisfy its equation.

Substituting two 𝑎 for 𝑥 and three 𝑎 for 𝑦 into the equation of the unit circle, we have that two 𝑎 all squared plus three 𝑎 all squared is equal to one. This gives an equation we can solve in order to find the value of 𝑎. Squaring each of the terms gives four 𝑎 squared plus nine 𝑎 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 𝑎 squared plus three 𝑎 squared is equal to one. This is an incorrect equation and would lead to an incorrect value of 𝑎.

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

Now this would suggest that there are two possible values for 𝑎: positive one over root 13 and negative one over root 13. But let’s check whether both of these are valid. Remember that 𝜃 is between zero and 𝜋 over two, which means that it’s situated in the first quadrant. This means that both of its coordinates are positive values. As two 𝑎 and three 𝑎 must, therefore, both be greater than zero, this means we need to take only the positive value of 𝑎. The value of 𝑎 is one over the square root of 13.