Where do you need help?

Each of these is a simply defined ratio taken from one of your "special" triangles.

The fact that you can't get even the simplest (#3, #6) shows that you haven't bothered to consult your textbook, your class notes, or a trivial web search; you haven't drawn and labeled the triangles. These are required by the guidelines before you post here.

That said, I expect that you're going to need the table I gave my trig students.

It contains the only angles you have to memorize for the quizzes & tests.

It's not in the simplest form, but it's easy to remember, and reducing the roots is simple

sin 00 = cos 90 = √0 / 2 = 0

sin 30 = cos 60 = √1 / 2 = 1/2

sin 45 = cos 45 = √2 / 2

sin 60 = cos 30 = √3 / 2

sin 90 = cos 00 = √4 / 2 = 1

See how easy? That column with the radicals is the critical one.

Now, for the other trig functions:

tan = sin / cos

cot = 1/tan

sec = 1/cos

csc = 1/sin

That and trivial algebra are all you need to know.

Can you do these now? If not, you'll have to explain where you need help.

Right triangles with a 30º angle or 30º, 60º, 90º , have opposite sides in

proportion to: 1 : √3 : 2

 tan[30º] = 1 ⁄ √3    cot[30º] = √3

 sin[30º] = 1 ⁄ 2     csc[30º] = 2

 cos[30º] = (√3) ⁄ 2    sec[30º] = 2 ⁄ √3

 tan[60º] = √3     cot[60º] = 1 ⁄ √3

 sin[60º] = (√3) ⁄ 2     csc60º] = 2 ⁄ √3

 cos[60º] = 1 ⁄ 2    sec[60º] = 2

Right triangles with a 30º angle or 45º, 45º, 90º , have opposite sides in

proportion to: 1 : 1 : √2

 tan[45º] = 1       cot[45º] = 1

 sin[45º] = 1 ⁄ √2    csc[45º] = √2

 cos[45º] = 1 ⁄ √2     sec[45º] = √2

1. Rectangular root of 48 + square root of three = ( 16 x three )^zero.5 + ( 3 ) ^zero.5 = 4 ( 3) zero.5 + 1 (three)^0.5 = 5 ( 3)^zero.5 or 5 x square root of three question 2 is not clear. 3. -four*sqaure root of ab *three rectangular root of b = -12 b square root of a.