# Thread: Trigonometry on the triangle rectangle

1. ## Trigonometry on the triangle rectangle

O being the center of the circle of radius unitáio, the area of triangle ABC has the catheter in diameter AC, it:

$\frac{3\sqrt{3}}{8}$

2. Originally Posted by Apprentice123
O being the center of the circle of radius unitáio, the area of triangle ABC has the catheter in diameter AC, it:

$\frac{3\sqrt{3}}{8}$
1. Unfortunately you forgot to tell us that $|\overline{OC}|$ is half of the radius. Since r = 1 you know that $|\overline{AC}| = \dfrac34$

2. The right triangle in question is part of the right triangle ADB (see attachment). The side CB is the height in the triangle ADB.

3. To calculate the length of the height use Euklid's theorem:

$|\overline{AC}| \cdot |\overline{CD}|= h^2~\implies~ h=\sqrt{\dfrac34 \cdot \dfrac14} = \dfrac14 \cdot \sqrt{3}$

4. Calculate the area of the triangle ACB:

$a = \dfrac12 \cdot |\overline{AC}| \cdot |\overline{CB}| = \dfrac12 \cdot \dfrac34 \cdot \dfrac14 \cdot \sqrt{3} = \dfrac3{16} \cdot \sqrt{3}$

According to your headline you maybe want to calculate the area of a rectangle(?). If so you have to double the result from #4. And then you'll get the result which is given by the answer.