How do I do this without a calculator

[Polonium 84]

I have used a calculator for this and got 192. Question how do I find these without a calculator is there a method that anyone knows?
[math]tan(30°) = \frac{ \sqrt{3}}{2}[/math]0
[math]tan(60^{\circ}) = \sqrt{3}[/math][math]12 × \frac{\sqrt{3}}{2} +2 × \sqrt{3} = 8\sqrt{3}[/math]

 

[Cubist]

Can you write 8 as the square root of something, say [imath]\sqrt{x}[/imath]? What is x?

EDIT: Actually, are you stuck on turning [imath]8\sqrt{3}[/imath] into [imath]\sqrt{k}[/imath] ? Or are you asking for an easy way to memorise the trig values for certain angles (@The Highlander might be able to help with that! See this post)

 

[topsquark]

View attachment 34097
I have used a calculator for this and got 192. Question how do I find these without a calculator is there a method that anyone knows?
[math]tan(30°) = \frac{ \sqrt{3}}{2}[/math]0
[math]tan(60^{\circ}) = \sqrt{3}[/math][math]12 × \frac{\sqrt{3}}{2} +2 × \sqrt{3} = 8\sqrt{3}[/math]

In most Introductory classes in Trigonometry the trig functions for 0, 30, 45, 60, and 90 degrees are usually taught. I'd look those up and memorize them, not because they have any real intrinsic value in themselves, but that many sources will assume you know them.

The trig values for these angles can be calculated by considering the properties of certain right triangles. See here.

-Dan

 

[The Highlander]

View attachment 34097
I have used a calculator for this and got 192. Question how do I find these without a calculator is there a method that anyone knows?
[math]tan(30°) = \frac{ \sqrt{3}}{2}[/math]0
[math]tan(60^{\circ}) = \sqrt{3}[/math][math]12 × \frac{\sqrt{3}}{2} +2 × \sqrt{3} = 8\sqrt{3}[/math]

Can you write 8 as the square root of something, say [imath]\sqrt{x}[/imath]? What is x?

EDIT: Actually, are you stuck on turning [imath]8\sqrt{3}[/imath] into [imath]\sqrt{k}[/imath] ? Or are you asking for an easy way to memorise the trig values for certain angles (@The Highlander might be able to help with that! See this post)

Thanks for the mention @Cubist ? but I don’t think the OP’s problem is in knowing the exact values for the 'usual suspects' (I have, however, included my summary sheet at the end; I’m a great believer in recycling?).

No, s/he has clearly been able to use the correct exact values, though I suspect that \(\displaystyle tan(30°) = \frac{\sqrt{3}}{2}\) is just a wee typo and s/he actually meant to type \(\displaystyle cos(30°) = \frac{\sqrt{3}}{2}\). ?

It would appear, rather, that the OP’s problem is a certain unfamiliarity with Simplifying Surds (or in this case, perhaps, Complicating Surds ?)

@Polonium 84: 192 is, of course, the correct answer for \(\displaystyle k\) but you need to study this web page and once you have learned what’s on there, this simple example should see you right:-

\(\displaystyle 10=\sqrt{100}\Rightarrow10\sqrt{5}=\sqrt{100}\sqrt{5}=\sqrt{100×5}=\sqrt{500}\)​

I Trust that will solve your 'problem'?

Here is the aforementioned Summary Sheet (that I believe @Cubist was referring to).

I've given you the sheet with all the "blanks" already filled in so all you need to do is make sure you know them all.
Enjoy.

 

[The Highlander]

Draw an equilateral triangle and draw a line from one vertex perpendicular to the opposite side. That line bisects the angle at the vertex as well as the opposite side. All three angles in an equilateral triangle are 180/3= 60 degrees. The new line divides it into two right triangles with one angle 60 degrees and the other 30 degrees.

Call the length of the sides of the equilateral triangle "s". Then each right triangle has hypotenuse of length s and one leg, opposite the 30 degree angle, s/2. The leg opposite the 60 degree angle is, by the Pythagorean theorem, sqrt(s^2- s^2/4)=sqrt(3s^2/4)= (s/2)sqrt(3)

The sine of 30 degrees is "opposite over hypotenuse"= (s/2)/s=1/2 and the cosine of 30 degrees is "near side over hypotenuse"= ((s/2) sqrt(3))/s= sqrt(3)/2.

For 60 degrees, since "near" and "opposite" sides are swapped, it is the reverse- the sine of 60 degrees is sqrt(3)/2 and the cosine is 1/2.

Just grab a copy of my Summary Sheet (above), @Shiloh, and the one below (for the users to fill in the blanks themselves).

That will save you an awful lot of typing (and minimize the risk of any little typing errors creeping into your "expressions") if you want to share that wisdom on Exact Values with anyone else. ??

​

All my work is "free" for anyone else to use, if desired and here are the BBCodes if they're more convenient:-

Without answers:  {img]https://i.ibb.co/kGqhDJ4/Exact-Values.png[/img]   (Usual one to give out)
With answers:     {img]https://i.ibb.co/3WwtZzh/Exact-Values-2.png[/img]

NB: You need to change the curly bracket at the start for a square one, of course. {→[