Question

Special Right Triangles? Can someone walk me through how to Find X and Y

I know one angle is 30, the other one is 90, so the last one must be 60. Does that have anything to do with solving x and y tho?

Answer 1

I have the answers i think but i have them in numbers, i need them in radicals..

Answer 2

use Hypotenuse * cos 30 = y and
Hypotenuse * sin 30 = x

If you had the other angle (60) then
use Hypotenuse * sin 60 = y and
Hypotenuse * cos 60 = x

Answer 3

Think of cosine as the hypotenuse "falling" onto the cosine side and when it stands back up, its rises the length of the sine side.

Answer 4

Haha, I'm lost on the cosine ans sine (I think there called that)

Answer 5

your hypotenuse determines both sides, it's just 20 in your graph

Answer 6

do you know how to compute cos(30) (i.e. cosine of 30 degrees)?

Answer 7

Not really, but I need to learn this haha

Answer 8

I don't want you to spoon feed me the answers, but do you mind helping me learn how to do it?

Answer 9

you press 30 in your calculator and they press the cos (cosine) function (make sure your calculator is set for degrees and not radians)

Answer 10

What do you get when you do this?

Answer 11

It says leave all answers as reduced radicals.

Answer 12

great,now multiply this with the hypotenus. The number sqrt(3)/2 is the fraction of the hypotenuse that lies on the opposite segment from the 30 degrees.

Answer 13

You can also think of hypotenuse as a the part of it that is horizontal (i.e. the cosine) and the part of it that is vertical (i.e. the sine)

Answer 14

\[20*\frac{ \sqrt{3} }{ 2 }=?\]

Answer 15

yes! That's "y". Now do the same thing for x.

Answer 16

except use sine, instead of cosine. Find sin(30)

Answer 17

yes. And you can check if this is right by using \[\sqrt{x ^{2}+y ^{2}}=20\]

Answer 18

Does this equal 20?
\[\sqrt{10^{2}+(10\sqrt{3})^{2}}\]

Answer 19

yes, that's your answer

Answer 20

How do I know which one to use cos and the other to use sin on?

Answer 21

If you have a triangle with the angle on the "ground", then the hypotenuse wants to fall on the ground towards the cosine part, that is, it wants to go horizontal. The sine part is the distance from the tip of the hypotenuse to the ground when it is "standing"