• Oct 10th 2008, 06:13 AM
Flexible
$5\sqrt{13}=a\sqrt{3}^2 + 10^2 - 2\times{a\sqrt{3}}\times{10}\times{cos150}$

How would i work out a? how do you square $a\sqrt3$ and $5\sqrt13$

Can someone please show me in stages how to fully expand $- 2\times{a\sqrt{3}}\times{10}\times{cos150}$ also as i'm not show how when surds and trigonometric functions are involved.

Many thanks guys!

F.
• Oct 10th 2008, 07:40 AM
masters
Quote:

Originally Posted by Flexible
$(5\sqrt{13})^2=(a\sqrt{3})^2 + 10^2 - 2a\sqrt{3}(10)\cos150$

How would i work out a? how do you square $a\sqrt3$ and $5\sqrt13$

Can someone please show me in stages how to fully expand $- 2\times{a\sqrt{3}}\times{10}\times{cos150}$ also as i'm not show how when surds and trigonometric functions are involved.

Many thanks guys!

F.

I assumed that you were using the law of cosines, so I tweaked your notation a little bit.

$(a\sqrt{3})^2=a^2 \cdot (\sqrt{3})^2=a^2 \cdot 3 = 3a^2$

$(5\sqrt{13})^2=5^2 \cdot (\sqrt{13})^2=25 \cdot 13 = 325$

Start here:

$325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)$
• Oct 10th 2008, 09:12 AM
Flexible
Thanks! i don't really know how to manipulate the surds? hence i cant expand and arrange to find a?
• Oct 10th 2008, 10:04 AM
masters
Quote:

Originally Posted by masters
I assumed that you were using the law of cosines, so I tweaked your notation a little bit.

$(a\sqrt{3})^2=a^2 \cdot (\sqrt{3})^2=a^2 \cdot 3 = 3a^2$

$(5\sqrt{13})^2=5^2 \cdot (\sqrt{13})^2=25 \cdot 13 = 325$

Start here:

$325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)$

By the way, $\cos 150=- \cos 30=-\frac{\sqrt{3}}{2}$

Now, to simplify:

$325=3a^2+100-20a\sqrt{3}\left(-\frac{\sqrt{3}}{2}\right)$

$325=3a^2+100-10a\sqrt{3}(-\sqrt{3})$

$325=3a^2+100+10a(3)$

$325=3a^2+100+30a$

$225=3a^2+30a$

$3a^2+30a-225=0$

$a^2+10a-75$

$(a+15)(a-5)=0$

$a+15 = 0 \ \ or \ \ a-5=0$

$a=-15 \ \ or \ \ a=5$

a=-15 is obviously an extraneous root, so a = 5 is your solution.
• Oct 10th 2008, 10:13 AM
Greengoblin
Quote:

Originally Posted by Flexible
Thanks! i don't really know how to manipulate the surds? hence i cant expand and arrange to find a?

The rule is $\sqrt{ab}=\sqrt{a}\sqrt{b}$
• Oct 10th 2008, 10:35 AM
Flexible
is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?
• Oct 10th 2008, 10:59 AM
masters
Quote:

Originally Posted by Flexible
is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?

You could start with the unit circle.
• Oct 10th 2008, 04:55 PM
Prove It
Quote:

Originally Posted by Flexible
is there anywhere i can find a formulae sheet with the trigonometric functions in surd form?

Yes, the Unit Circle helps, but it also helps to have the special triangles on hand so you can find the trig functions of the focus angles.

Draw a right angled isosceles triangle with the identical sides = 1 unit in length. Clearly its angles are $45^\circ, 45^\circ, 90^\circ$.

By Pythagoras, the hypotenuse is $\sqrt{2}$ units in length.

What are $\sin{45^\circ}, \cos{45^\circ}$ and $\tan{45^{\circ}}$?

Then draw an equilateral triangle of side length 2 units. Draw a perpendicular bisector, then you have two $30^\circ, 60^\circ, 90^\circ$ triangles. Their side lengths are $1, \sqrt{3}, 2$ units (check using Pythagoras).

What are $\sin{30^\circ}, \cos{30^\circ}, \tan{30^\circ}, \sin{60^\circ}, \cos{60^\circ}$ and $\tan{60^\circ}$?