Two corners of a triangle have angles of $\frac{2 π}{3}$ $(2 pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $15$ $15$ , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{2 π}{3}$ $(2 pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $15$ $15$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-27T04:54:18

$P = 106.17$ $P = 106.17$

Explanation:

By observation, the longest length would be opposite the widest angle, and the shortest length opposite the smallest angle. The smallest angle, given the two stated, is $\frac{1}{12} (π)$ $1/12(pi)$ , or $15^{o}$ $15^o$ .

Using the length of 15 as the shortest side, the angles on each side of it are those given. We can calculate the triangle height $h$ $h$ from those values, and then use that as a side for the two triangular parts to find the other two sides of the original triangle.
$tan (\frac{2}{3} π) = \frac{h}{15 - x}$ $tan(2/3pi) = h/(15-x)$ ; $tan (\frac{1}{4} π) = \frac{h}{x}$ $tan(1/4pi) = h/x$

$- 1.732 = \frac{h}{15 - x}$ $-1.732 = h/(15-x)$ ; $1 = \frac{h}{x}$ $1 = h/x$
$- 1.732 \times (15 - x) = h$ $-1.732 xx (15-x) = h$ ; AND $x = h$ $x = h$ Substitute this for x:

$- 1.732 \times (15 - h) = h$ $-1.732 xx (15-h) = h$
$- 25.98 + 1.732 h = h$ $-25.98 + 1.732h = h$

$0.732 h = 25.98$ $0.732h = 25.98$ ; $h = 35.49$ $h = 35.49$
Now, the other sides are:
$A = \frac{35.49}{sin (\frac{π}{4})}$ $A = 35.49/(sin(pi/4))$ and $B = \frac{35.49}{sin (\frac{2}{3} π)}$ $B = 35.49/(sin(2/3pi))$

$A = 50.19$ $A = 50.19$ and $B = 40.98$ $B = 40.98$

Thus, the maximum perimeter is:
$P = 15 + 40.98 + 50.19 = 106.17$ $P = 15 + 40.98 + 50.19 = 106.17$

Answer 2 · 2017-12-27T05:23:31

Perimeter $= 106.17$ $=106.17$

Explanation:

let
$∠ A = \frac{2 π}{3}$ $angle A=(2pi)/3$
$∠ B = \frac{π}{4}$ $angle B=pi/4$
therefore;
using angle sum property
$∠ C = \frac{π}{12}$ $angle C=pi/12$

Using the sine rule

![https://www.youtube.com/watch?v=bDPRWJdVzfs]( useruploads.socratic.org )

$a = 15 \times \frac{sin (\frac{2 π}{3})}{sin (\frac{π}{12})} = 50.19$ $a=15×sin ((2pi)/3)/sin (pi/12) = 50.19$
$b = 15 \times \frac{sin (\frac{π}{4})}{sin (\frac{π}{12})} = 40.98$ $b=15×(sin ((pi)/4))/sin (pi/12) = 40.98$

perimeter $= 40.98 + 50.19 + 15 = 106.17$ $=40.98+50.19+15 =106.17$

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Two corners of a triangle have angles of $\frac{2 π}{3}$ $(2 pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $15$ $15$ , what is the longest possible perimeter of the triangle?

2 Answers

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Two corners of a triangle have angles of (2 pi )/ 3 2π3 (2 pi )/ 3 and ( pi ) / 4 π4 ( pi ) / 4 . If one side of the triangle has a length of 15 15 15 , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{2 π}{3}$ $(2 pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $15$ $15$ , what is the longest possible perimeter of the triangle?