TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg
I worked out that it would be 127 cm2
TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg
I worked out that it would be 127 cm2
No.
For any triangle, it's area can be found using the formula
$\displaystyle A = \frac{1}{2}\,ab\,\sin{C}$, where $\displaystyle C$ is the angle in between sides $\displaystyle a$ and $\displaystyle b$.
So $\displaystyle A = \frac{1}{2}\cdot 12\,\textrm{m}\cdot 30\,\textrm{m} \cdot \sin{45^{\circ}}$
$\displaystyle = 180\,\textrm{m}^2\cdot \frac{\sqrt{2}}{2}$
$\displaystyle = 90\sqrt{2}\,\textrm{m}^2$.
That's exactly the same rule. I've just chosen two different sides of the triangle.
Point is, you need two sides of the triangle and the angle between them.
BTW I always like to keep the solutions exact where possible, hence the square root value rather than the decimal.
I'm not sure why you reposted this question when I already answered it in your other thread: http://www.mathhelpforum.com/math-he...-triangle.html. Did you not see it?
01
Like I said, I've kept the answer exact.
The answer is not $\displaystyle 90$.
The answer is $\displaystyle 90 \times \sqrt{2}$.
Notice that when you put it into the calculator you get a decimal answer. This is NOT exact, it is a decimal approximation to $\displaystyle 90\times\sqrt{2}$.
You have done the right thing but have gotten the decimal answer rather than the exact answer.