A triangle has two corners with angles of $\frac{π}{4}$ $( pi ) / 4$ and $\frac{7 π}{12}$ $( 7 pi )/ 12$ . If one side of the triangle has a length of $8$ $8$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-05-28T18:15:20

$A = 32 \sqrt{2} sin (\frac{7 π}{12})$ $A = 32sqrt2sin((7pi)/12)$

Find the third angle:

$π - \frac{π}{4} - \frac{7 π}{12} = \frac{π}{6}$ $pi-pi/4-(7pi)/12 = pi/6$

We will get the largest triangle, if we make the smallest angle be opposite the given side:

Let $∠ A = \frac{π}{6}$ $angle A = pi/6$ and side $a = 8$ $a = 8$

Let $∠ B = \frac{π}{4}$ $angle B = pi/4$

Let $∠ C = \frac{7 π}{12}$ $angle C = (7pi)/12$

Use the Law of Sines to find the length of side b:

$\frac{b}{sin (B)} = \frac{a}{sin (A)}$ $b/sin(B) = a/sin(A)$

$b = \frac{a}{sin (A)} sin (B)$ $b = a/sin(A)sin(B)$

$b = \frac{8}{sin (\frac{π}{6})} sin (\frac{π}{4})$ $b = 8/sin(pi/6)sin(pi/4)$

$b = \frac{8}{0.5} \frac{\sqrt{2}}{2}$ $b = 8/0.5sqrt(2)/2$

$b = 8 \sqrt{2}$ $b = 8sqrt2$

The height is:

$h = b sin (C)$ $h = bsin(C)$

$h = 8 \sqrt{2} sin (\frac{7 π}{12})$ $h = 8sqrt2sin((7pi)/12)$

The Area is:

$A = \frac{1}{2} base \times height$ $A = 1/2"base"xx"height"$

We let side a be the base:

$A = \frac{1}{2} (8) (8 \sqrt{2} sin (\frac{7 π}{12}))$ $A = 1/2(8)(8sqrt2sin((7pi)/12))$

$A = 32 \sqrt{2} sin (\frac{7 π}{12})$ $A = 32sqrt2sin((7pi)/12)$

A triangle has two corners with angles of ( pi ) / 4 π4 ( pi ) / 4 and ( 7 pi )/ 12 7π12 ( 7 pi )/ 12 . If one side of the triangle has a length of 8 88 , what is the largest possible area of the triangle?