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

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Answer 1 · 2017-07-13T12:40:48

The area of the triangle is $=58.2u^2$

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The angles aare

$hat(C)=1/12pi$

$hat(A)=1/3pi$

The third angle of the triangle is

$hat(B)=pi-(1/12pi+1/3pi)$

$=pi-5/12pi$

$=7/12pi$

Let the length of the side $c=6$ which is opposite the smallest angle of the triangle.

Then, we apply the sine rule to the triangle

$b/sin(7/12pi)=a/sin(pi/3)=6/sin(1/12pi)=23.18$

$b=23.18*sin(7/12pi)=22.39$

$a=23.18*sin(pi/3)=20.07$

The area of the triangle is

$=1/2ab sin(hat(C))=1/2*20.07*22.39*sin(pi/12)$

$=58.2$

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