On this type of problem, we're trying to figure out what five times the secanopy thirds +7 times the tangent of π thirds. Well, the first thing that we need to recall or understand is that π thirds is really the same thing as 60° because if I multiply by 180 / π° over radians, the Pi is canceled. Three goes into 180 sixty degrees. So when I see secant of π thirds, that's really saying secant of 60 or tangent of π thirds is really saying tangent of 60. If we think about our unit circle and we put in an angle of π thirds, which remember is really just 60, I put in this angle here and call it π / 3. Then if I drop an altitude straight down, that's going to form a right triangle, and this is really a 3060 right triangle because the angle up here is π six, which is equivalent to 30°. Well, in a unit circle we know that the hypotenuse has to be one. We also know that if we had drawn in the other half of this equilateral triangle, this whole base here would be a distance of one. So when we drop the altitude, we split this angle in half, which means we split the side in half so this side becomes 1/2. So now I need to figure out what the height is of this triangle. Well, 1/2 ^2 + H ^2 would equal 1 ^2 by Pythagorean theorem, so 1/2 ^2 is 1/4 + H ^2 equaling one. If I subtract the one fourth one is really 4 / 4, so 4 / 4 -, 1/4 is 3/4. Then if I take the square root, I'm going to think of it being positive or negative, and that's going to depend on which quadrant we're in root 3 / 2. So this point right here has an X value of 1/2. We went over 1/2, we went up root 3 / 2. So now the secant of π thirds is really the same thing as one divided by the cosine of π thirds. Well, cosine in our unit circle is just our X value, so it's going to be 1 / 1/2, or in this case just the number 2. Now we're going to do the same thing for our tangent. So our tangent of π thirds by definition is sine of π thirds divided by cosine of π thirds. If I have sine of π thirds divided by cosine of π thirds, the sine or the height was root 3 / 2. The cosine was 1/2. So the tangent of π thirds is just root 3. So now that we know the secant and the tangent, we can go back to the original and actually solve it. So we're going to have five times secant of π thirds. Well, secant of π thirds was 2 + 7 times the tangent of π thirds. The tangent of π thirds is root 3. So our final answer should be 10 + 7 square roots of three.