Two corners of a triangle have angles of $\frac{π}{6}$ $pi / 6$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{π}{6}$ $pi / 6$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-01-10T10:26:59

45.314cm

Explanation:

The three angles for triangle are $\frac{π}{6}, \frac{π}{12} and \frac{3}{4} π$ $pi/6, pi/12 and 3/4pi$

To get the longest perimeter, the shortest length shall reflex to the smallest angle.
Let say that the other lengths are b reflex to angle $\frac{π}{6}$ $pi/6$ and c reflex to angle $\frac{3}{4} π$ $3/4pi$ while a=8 reflex to angle $\frac{π}{12}$ $pi/12$

therefore
$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA =b/sinB =c/sinC$

$\frac{b}{sin (\frac{π}{6})} = \frac{8}{sin (\frac{π}{12})}$ $b/sin(pi/6) =8/sin(pi/12)$

$b = \frac{8}{sin (\frac{π}{12})} \cdot sin (\frac{π}{6})$ $b=8/sin(pi/12)*sin(pi/6)$
$b = \frac{8}{0.2588} \cdot 0.5$ $b=8/0.2588*0.5$
$b = 15.456$ $b=15.456$

$\frac{c}{sin (\frac{3 π}{4})} = \frac{8}{sin (\frac{π}{12})}$ $c/sin((3pi)/4) =8/sin(pi/12)$

$c = \frac{8}{sin (\frac{π}{12})} \cdot sin (\frac{3 π}{4})$ $c=8/sin(pi/12)*sin((3pi)/4)$
$c = \frac{8}{0.2588} \cdot 0.7071$ $c=8/0.2588*0.7071$
$c = 21.858$ $c=21.858$

The longest possible perimeter = a+b+c
$= 8 + 15.456 + 21.858$ $=8+15.456+21.858$
$= 45.314 c m$ $= 45.314cm$

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Two corners of a triangle have angles of $\frac{π}{6}$ $pi / 6$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Two corners of a triangle have angles of $\frac{π}{6}$ $pi / 6$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?