How do you do this math problem: special right triangles?

Question

use special right triangles to find the coordinates of the point of intersection of the angle 150 degrees and the unit circle. Express the answer in fractions and radical when necessary. The answer is -rad3/2, 1/2...But HOW DO you get to that point
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Answer 1

If you start at where 3 would be on a clock and go counter-clockwise 150 degrees, you get to 30 degrees above the x axis. The radius, the x axis and the height between the x axis and where the ray touches the circle form a special 30-60-90 right triangle. Since the reference angle is 30 degrees, the opposite side, or the y coordinate, must be 1/2 (since the length of the 30 degree side is always 1/2 that of the 90 degree side and in a unit circle the 90 degree side is always 1). By that same rule, the x axis length (x coordinate) must be -1/2 root 3 because it is in the 2nd quadrant, making it negative, and it is the length of the 30 degree side times root three, since it is the 60 degree side.

Hope this helped!

Answer 2

Ok I'll try my best. The lengths of the sides of a 30-60-90 triangle are 1 (cooresponds with 90), 1/2 (cooresponds with 30), and Rad 3/2 (cooresponds with 60). So on a unit circle of 360 deg, you could draw a line representing 150 deg., and then straight down from there, forming a 30-60-90 traingle in Quadrant II. Then all you have to do is write in the degree values and side lengths on your traingle. the x-coordinate on 150 degrees would be across from the 60 deg, and since the 60 deg side length is Rad 3/2, and you are in the negatives (Quadrant II) The x-coordinate would be -Rad 3/2. And then its y-coordinate would be across from 30 deg, meaning it is 1/2. And since you go up it is positive.

Try drawing and writing down some of what I said. Hope I was a help!

Answer 3

sounds like pre-cal. let's see. that looks like a 30, 60 90, triangle . something with that. get the points on the side i guess. (negative radial 3, 1/2) yea. that's the answer. look at the unit circle.

Answer 4

a particular triangle has equivalent angles yet therefore there are purely 20, 70, and ninety, which for this reason, excludes it from being called a particular triangle. This math problem will be solved through utilising one in each of numerous trigonometric guidelines called the regulation of sines or the regulation of cosines. The regulation of sines entail that given sides s1, s2, and s3 alongside with their linked angles, the triangle comply to top the following criteria: ASA, SAS, or AAS yet no longer SSS or SSA (2 sides with an lined attitude). For this triangular sources, the angles 20, 70 have sides defined for SideA && SideC yet no longer for aspect B. for this reason, the triangle meets the 0.33 case with both angles defined for aspect A and SideC yet no longer for SideB. you should use the regulation of sines to derive both remnant sides inclusive of the attitude for aspect B through c / SinC = a / SinA || c / SinC = a / SinA. Likwise, understanding that ninety ranges - 20 ranges 70 ranges, the regulation of sines facilitates both the unknown attitude and aspect for B to be received through b / SinB = a / SinA. J.C

Answer 5

look at your text book and try to study before asking for the answer