A triangle has two corners with angles of $\frac{2 π}{3}$ $(2 pi ) / 3$ and $\frac{π}{6}$ $( pi )/ 6$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?

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Answer 1 · 2016-11-16T17:35:06

$A r e a = 4 \sqrt{3}$ $Area = 4sqrt(3)$

Let $∠ A = \frac{π}{6}$ $angle A = pi/6$

Let $∠ B = \frac{2 π}{3}$ $angle B = (2pi)/3$

Then $∠ C = π - ∠ A - ∠ B$ $angle C = pi - angle A - angle B$

$∠ C = \frac{π}{6}$ $angle C = pi/6$

Let side $a = 4$ $a = 4$

Because $∠ A = ∠ C$ $angle A = angle C$ , then side $c = 4$ $c = 4$ , too.

We can use the Law of Cosines to find the length of side b:

$b = \sqrt{a^{2} + c^{2} - 2 (a) (c) cos (B)}$ $b = sqrt(a^2 + c^2 - 2(a)(c)cos(B))$

$b = \sqrt{4^{2} + 4^{2} - 2 (4) (4) cos (\frac{2 π}{3})}$ $b = sqrt(4^2 + 4^2 - 2(4)(4)cos((2pi)/3))$

$b = 4 \sqrt{3}$ $b = 4sqrt(3)$

Let b = the base of the triangle

Let h = the height of the triangle

$h = a sin (A)$ $h = asin(A)$

$h = 4 sin (\frac{π}{6})$ $h = 4sin(pi/6)$

$h = 2$ $h = 2$

$A r e a = (\frac{1}{2}) b h$ $Area = (1/2)bh$

$A r e a = (\frac{1}{2}) (4 \sqrt{3}) (2)$ $Area = (1/2)(4sqrt(3))(2)$

$A r e a = 4 \sqrt{3}$ $Area = 4sqrt(3)$

A triangle has two corners with angles of (2 pi ) / 3 2π3 (2 pi ) / 3 and ( pi )/ 6 π6 ( pi )/ 6 . If one side of the triangle has a length of 4 44 , what is the largest possible area of the triangle?