# Solving problem: Area of Triangle

I have this data:

• $a=6$

• $b=3\sqrt2 -\sqrt6$

• $\alpha = 120°$

How to calculate the area of this triangle?

there is picture:

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What area formulae do you know about triangles? –  Sean Jun 6 '12 at 17:13
i know all formuleas abouth triangles –  Luka Toni Jun 6 '12 at 17:15
@Luka Wow! That's impressive. –  Rick Decker Jun 6 '12 at 17:19
@LukaToni Its next to impossible. –  Sawarnik Nov 13 at 4:01

Because the angle at $A$ is obtuse, the given information uniquely determines a triangle. To find the area of a triangle, we might want:

• the length of a side and the length of the altitude to that side (we don't have altitudes)
• all three side lengths (we're short 1)
• two side lengths and the measure of the angle between them (we don't have the other side that includes the known angle or the angle between the known sides)

(There are other ways to find the area of a triangle, but the three that use the above information are perhaps the most common.)

Let's find the angle between the known sides (since we'd end up finding that angle anyway if we were trying to find the unknown side). The Law of Sines tells us that $\frac{\sin A}{a}=\frac{\sin B}{b}$, so $$\frac{\sin120^\circ}{6}=\frac{\sin B}{3\sqrt{2}-\sqrt{6}},$$ which can be solved for $B$ (since $A$ is obtuse, $0^\circ<B<90^\circ$, so there is a unique solution). Once we have $B$, we can use $A+B+C=180^\circ$ to get $C$ and then the area of the triangle is $$\frac{1}{2}ab\sin C.$$

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Thanks for equation and super explanation :) –  Luka Toni Jun 6 '12 at 19:53

Assuming the diagram like so, where C = $\alpha=120$

Then we have the equation

$$Area = \displaystyle\frac{a b\sin C}{2}$$

This is the same as the equation you probably know,

$$Area = \displaystyle\frac{base*height}{2}$$

Do you know why?

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How about using the cosine rule? –  Sean Jun 6 '12 at 17:26
shrani.si/f/2Q/MU/3fecXbig/sss.jpg there is picture with A, B , C points –  Luka Toni Jun 6 '12 at 17:29