Trigonometry help

Freqm0n

I am going through the motions over at Khan Academy and working my way through some very basic math which escaped me in my younger years. I have searched and watched all material that I believe is correct to the subject but some thing just won't click for me.

30-60-90 Triangles

All I am given is a triangle with a hypotenuse length of 6√​3 and asked to find X (the long side, or side opp. the 60​∘ angle)

so, the short side (or side opp. the 30​∘) angle must be half 6√​3
=3√​3 .. which is simply √​3 <--- is this correct?

If so, X = √​3 times √​3

I'm unsure at this point if this is correct, or even the semantics of approaching the calculation "√​3 times √​3" to get the answer in the format they want.

The Question is looking for an answer in the format X√​X

Any help greatly appriciated

TheBody

Hi there!
If I am reading your problem correctly you could do the following.

Using the trig ratio, [latex]\sin(60)=\frac{x}{6\sqrt{3}}[/latex].

Transposing for x we get: [latex]x=6\sqrt{3}\sin(60)=9[/latex].

Freqm0n

I don't get it

It stated in the material that to find the long side we must multiply the short side by √​3

...I don't know the short side, but this can be got by halving the hypotenuse which is 6√​3 = √​3

so to find the long side: √​3.√​3 = ? ... (in the format X√​X)

TheBody

Do you know the trig ratios? sin cos and tan....

http://www.mathsisfun.com/algebra/trigonometry.html

TimeToShine

Freqm0n wrote: »

I am going through the motions over at Khan Academy and working my way through some very basic math which escaped me in my younger years. I have searched and watched all material that I believe is correct to the subject but some thing just won't click for me.

30-60-90 Triangles

So it's a right angled triangle.

All I am given is a triangle with a hypotenuse length of 6√​3 and asked to find X (the long side, or side opp. the 60​∘ angle)

You are given the hypotenuse length and the three angles.

so, the short side (or side opp. the 30​∘) angle must be half 6√​3
=3√​3 .. which is simply √​3 <--- is this correct?

No. I don't know where you got this from but it is completely wrong.

edit: You aren't technically wrong but you seem to have learned this off for this particular triangle which isn't the right way to go about it. See below.

The sine rule is, provided you are given a side and the two opposite angles - which you have here.

a/sin(a) = b/sin(b) = c/sin(c)

So

sin(90)/6√​3 = sin(30)/x = sin(60)/y

you want x so:

sin(90)/6√​3 =sin(30)/x

So

sin(90)/6√3 = 0.5/x

x = 3√​3

TheBody

TimeToShine wrote: »

The sine rule is, provided you are given a side and the two opposite angles - which you have here.

a/sin(a) = b/sin(b) = c/sin(c)

So

sin(90)/6√​3 = sin(30)/x = sin(60)/y

you want x so:

sin(90)/6√​3 =sin(30)/x

So

sin(90)/6√3 = 0.5/x

x = 3√​3

I don't think you are reading the set up correctly. He says that the side x is opposite the [latex]60^\circ [/latex] angle.

At any rate, the sine rule is a little bit of overkill.

TimeToShine

TheBody wrote: »

I don't think you are reading the set up correctly. He says that the side x is opposite the [latex]60^\circ [/latex] angle.

At any rate, the sine rule is a little bit of overkill.

Yeah you're right my bad.

Pherekydes

Freqm0n wrote: »

I don't get it

It stated in the material that to find the long side we must multiply the short side by √​3

...I don't know the short side, but this can be got by halving the hypotenuse which is 6√​3 = √​3

so to find the long side: √​3.√​3 = ? ... (in the format X√​X)

Half the hypotenuse is 3√3, which is definitely not √3. This error is probably throwing you off.

mickdw

All you need when dealing with right angle triangle and the info you have is pythagoros
So sloped side is 6 root3
Short side is half that so 3 root3 (equilateral triangle)
Call unknown side X
So Using pythagoros:

108 = 27 + X squared
X squared = 108-27
X squared = 81
X =9

Or in required format:

3 root9.

...............................................

I cannot write the correct symbols here so couldn't write out the calculation fully but just to explain the above

The 108 comes from 6 root3 squared
The 27 comes from 3 root3 squared.