I would like to calculate the length of the side in red on the image. I tried the Law of cosines, but maybe i haven't applied the formula right, because for a side "a" and "b" of size 64 and a angle of 120, the result is 39.
How to calculate the right length of c for the image?
3 Answers
$\begingroup$In an isosceles triangle as shown, the bisector of the $120^\circ$ angle is the perpendicular bisector of the base. So $$\frac c2=b\sin60^\circ\ .$$
$\endgroup$ $\begingroup$Split the triangle in two right triangles whose lower or base side is $\frac{c}{2}.\;$ Now these "new triangles" have angles, $60^\circ$, $30^\circ$ and $90^\circ$, respectively.
As $\cos(30^\circ) = \frac{\sqrt{3}}{2} $, now you can easily check that $\frac{c}{2} = \frac{64 \cdot \sqrt{3}}{2} = 32 \cdot \sqrt{3}$. So $c = 64 \cdot \sqrt{3}$.
As for the other side, you must do the same thing but with sine. $\sin(30^\circ) = \frac{1}{2}$, so the height of the triangle is $32$.
$\endgroup$ 2 $\begingroup$First you can bisect your angle given ($120$ degrees) to $60$ degrees. Then you will notice that you will have formed two triangles that follow the $30$-$60$-$90$ triangle format. $64$ is your hypotenuse and the triangle side you're looking for is $64\sqrt{3}$.
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