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Hi I need to figure out the area of the following triangle, without using Trigonometric ratios. Any suggestions on the best approach.

enter image description here

The answer is 12 square units

Edit: I also think that the above triangle can't qualify for a $30-60-90$ triangle since it fails the $x,x.\sqrt3,2x$ rule/

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    (cont...) Some of your edits might be rejected and you will not be able to [suggest edits](http://meta.math.stackexchange.com/questions/4672/).2012-12-07

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If you use a 30-60-90 triangle with hypotenuse 6, then the height is 3. So the height of this triangle is 3. Thus $\frac{b\cdot h}{2}=\frac{8\cdot3}{2}=12$

triangle

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    Isn't knowing the ratios of the sides of a triangle with given angles essentially where trigonometry begins? (+1 for the answer, in any case).2012-06-17
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This can be done without using any trigonometric functions. Let $|AC| = 6$, $|BC| = 8$ and $|\angle ACB| = 30^\circ$. Let $H \in BC$ be a point such that $HA$ is the height of the triangle starting at $A$. Then, the $\triangle CAH$ is a half of equilateral triangle and therefore $|HA| = 3$. Using the basic formula for the triangle's area we get $\frac{|CB|\cdot|HA|}{2} = \frac{8\cdot 3}{2} = 12$.

Edit: Considering your last modifications to the question, please take look at the picture below. Let $G$ be a point on the line passing through $A$ and $H$ such that $|GH|=|AH|$ and $|AG|=2|GH|$. Please note that $\triangle AGC$ is now equilateral and thus $\triangle AHC$ is the 30-60-90 triangle.

enter image description here

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    @MistyD No. Let $G$ be a point on the line passing through $A$ and $H$ such that $|GH| = |AH|$ and $|AG| = 2|GH|$. Then $\triangle AGC$ is equilateral, so $|AG| = |AC| = 6$ and $|AH| = 3$. This the same solution as Joseph's in different wording, he was just faster ;-)2012-06-17
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You can find the area of any triangle by applying the following area formula:

$A=\frac{1}{2}ab\sin(C),$

Where $a$ and $b$ are sides of the triangle and $C$ is the angle between them.

In this case, you can do the following:

$\frac{6\times8}{2}\sin(30^{\circ})=12$

Which is the answer you want.

EDIT: In response to your comment, you can find more about this formula and it's derivation, here

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    @dtldarek I interpreted that as him wishing to stay clear of right-triangle specific trigonometric ratios: $\sin(\theta)=\frac{o}{h}$ for instance, rather than trigonometric functions in general.2012-06-17