1. ## Geometry Trouble

Need urgent help!
Using the technique in the model above, find the missing sides in this 30°-60°-90° right triangle. Hypotenuse = 12
Long=?

2. ## Re: Geometry Trouble

$6\sqrt{3}$?

3. ## Re: Geometry Trouble

The first portion says "Using the technique in the model above" but you didn't include that in your post!

Lacking that here is what I might do: An equilateral triangle has all three sides of the same length and all three angles the same measure. That common length could be anything but since the angles in a triangle must add to 180 degrees, each angle has measure 180/3= 60 degrees. Now, imagine drawing a line from one vertex to the center of the opposite side. By symmetry, the two angles formed at the vertex must be the same so 60/2= 30 degrees and the angle form at the side must be the same: 180/2= 90 degrees. That is, we have formed to right triangles with angles 60 and 30 degrees and one leg with length half the length of the hypotenuse. Here, that hypotenuse has length 12 so one leg has length half that, 6. Use the Pythagorean theorem to determine the length of the other leg.

4. ## Re: Geometry Trouble

By using the definition of sine and cosine! Given angle $\displaystyle \theta$ in a right triangle, $\displaystyle sin(\theta)$ is defined as "opposite side divided by hypotenuse". Here the "long side" is opposite the 60 degree angle and the hypotenuse is given as 12. So, calling the length of the long side "x", $\displaystyle \frac{x}{12}= sin(60)$ so $\displaystyle x= 12 sin(60)$.
Equivalently, $\displaystyle cos(\theta)$ is defined as "near side divided by hypotenuse". Here the "long side"is "near" the 30 degree angle and the hypotenuse is given as 12. So, calling the length of the long side "x" again, $\displaystyle \frac{x}{12}= cos(30)$ so $\displaystyle x= 12 cos(30)$.