Hello Forum.

I'm having trouble with this problem. Hope this is the right section, and that I translate it right the right way, so you can understand...

We have a right triangle. Smallest leg - AB. Biggest leg - BC. Hypotenuse - CA. We know that the hypotenuse has 10 centimeters, and that x is the angle BCA. How do I prove that the area of this triangle is given by A(x) = 25 sin (2x).

Any help would be appreciated. Thank you.

Since it's a right angle triangle

\(\displaystyle A(x) = \frac{b\,h}{2}\)

\(\displaystyle = \frac{(AB)(BC)}{2}\).

Recall that \(\displaystyle \sin{2x} = 2\sin{x}\cos{x}\).

Also recall that \(\displaystyle \sin{x} = \frac{\textrm{Opposite}}{\textrm{Hypotenuse}}\) and \(\displaystyle \cos{x} = \frac{\textrm{Adjacent}}{\textrm{Hypotenuse}}\).

So \(\displaystyle \sin{x} = \frac{AB}{10}\) and \(\displaystyle \cos{x} = \frac{BC}{10}\).

Therefore \(\displaystyle \sin{2x} = 2\sin{x}\cos{x}\)

\(\displaystyle = 2\left(\frac{AB}{10}\right)\left(\frac{BC}{10}\right)\)

\(\displaystyle = \frac{(AB)(BC)}{50}\).

Therefore \(\displaystyle 25\sin{2x} = \frac{(AB)(BC)}{2}\)

\(\displaystyle 25\sin{2x} = A(x)\).