# A triangle has two corners with angles of  (2 pi ) / 3  and  ( pi )/ 6 . If one side of the triangle has a length of 12 , what is the largest possible area of the triangle?

May 23, 2018

$36 \sqrt{3}$

#### Explanation:

If we convert the two angle measures into degrees, we are given:

$\frac{2 \pi}{3} \cdot \frac{180}{\pi} = 120$

$\frac{\pi}{6} \cdot \frac{189}{\pi} = 30$

Based on the fact that triangle's angles must add up to 180 degrees, the third angle must be 30 degrees.

Our triangle's angles are 30, 30, and 120. This means that we have an isosceles triangle. It also means that we can get two different areas based on which side is 12.

Option 1: If the side with a length of 12 is opposite either 30 degree angle, then by the isosceles triangle theorem, the side opposite the other 30 degree angle must also be 12.

We'll use this triangle to illustrate our problem, with sides AB and BC being the two sides with lengths of 12, and BC as the side opposite the 120 degree angle, B. If we construct an altitude in the triangle as shown, we create two congruent (proven by the hypotenuse-leg theorem), 30-60-90 right triangles.

Using the 30-60-90 theorem, we know that side AB is double the length of side BM. Since AB is 12, BM is thus 6.

We also know (based on the said theorem) that side AM is the length of BM, multiplied by $\sqrt{3}$ (it's also half the length of side AC). Thus, it is $6 \sqrt{3}$. From this we can also say that AC is $12 \sqrt{3}$

With these measurements, we now have a base (side AC, $12 \sqrt{3}$), and a height (side BM, 6). We can use these values to find the triangle's area.

$12 \sqrt{3} \cdot 6 \cdot \frac{1}{2} = 36 \sqrt{3}$ This is the area of the triangle given that one the congruent legs is 12.

Option 2: If the side opposite the 120 degree angle is 12, we can apply the same logic as we did in option one. This time however, AC is 12. Once more using the 30-60-90 and congruency rules:

AM = MC= 6

BM * $\sqrt{3}$ = AM

BM = $\frac{6}{\sqrt{3}}$

Since we already have our base (side AC, 12), and height
(BM, $\frac{6}{\sqrt{3}}$), we can again find this triangle's area.

$12 \cdot \frac{6}{\sqrt{3}} \cdot .5 = \frac{36}{\sqrt{3}}$

$36 \sqrt{3} > \frac{36}{\sqrt{3}}$, so the largest possible area is $36 \sqrt{3}$.