Proving length of triangle (3D trig)

[Monkeyseat]

Sorry to post it in picture form but it's hard to show this.

Yeah I know I've posted a couple of questions, apologies, I have an exam tomorrow and I just want to find out how to do the ones I can't do.

Cheers.

 

[morson]

Very straightforward. Use Pythagoras' theorem in both questions.

 

[Monkeyseat]

I have:

a^2 + b^2 = c^2

3^2 + b^2 = 6^2

b^2 = 6^2 - 3^2

b^2 = 36 - 9

b^2 = 25

b=5

I get that but I don't know how to get 3sqr.3?

Please help lol.

 

[morson]

Since when is 36 - 9 equal to 25?!

 

[Monkeyseat]

Since when is 36 - 9 equal to 25?!

Hahaha. Sorry about that.

Yeah I've done A now, but for B, Im just wondering, how do you multiply 3sqr.3 by 2/3? Is it 2sqr.3? I know the method to do B, but I'm not really good at multiplying surds.

I know it's something like:

(2sqr.3)^2 (if that's right) + b^2 = 6^2

12 + b^2 = 36

b^2 = 24

b = sqr.4 * sqr.6

b = 2sqr.6

Am I close? :wink: Please help.

Thanks.

 

[jonboy]

Yeah I've done A now, but for B, Im just wondering, how do you multiply 3sqr.3 by 2/3? Is it 2sqr.3? I know the method to do B, but I'm not really good at multiplying surds.

\(\displaystyle \L \;3\sqrt[3]{3}\,\cdot\,\frac{2}{3}\,=\,(3\,\cdot\,\frac{2}{3})\sqrt[3]{3}\,=\,2\sqrt[3]{3}\)

 

[Monkeyseat]

Yeah I've done A now, but for B, Im just wondering, how do you multiply 3sqr.3 by 2/3? Is it 2sqr.3? I know the method to do B, but I'm not really good at multiplying surds.

\(\displaystyle \L \;3\sqrt[3]{3}\,\cdot\,\frac{2}{3}\,=\,(3\,\cdot\,\frac{2}{3})\sqrt[3]{3}\,=\,2\sqrt[3]{3}\)

It's not cube root, it's square root isn't it?

Is my above method correct?

TM is \(\displaystyle 2sqrt{6}\) ?

 

[jonboy]

Is my above method correct?

Whoops I confused myself. \(\displaystyle \;3\sqrt{3}\,\cdot\,\frac{2}{3}\,=\,2\sqrt{3}\).

Good job.

 

[Monkeyseat]

Is my above method correct?

Whoops I confused myself. \(\displaystyle \;3\sqrt{3}\,\cdot\,\frac{2}{3}\,=\,2\sqrt{3}\).

Good job.

No worries, can you confirm wether my final answer of TM is \(\displaystyle 2\sqrt{6}\) is correct?

Thanks.