Two corners of a triangle have angles of $\frac{3 π}{8}$ $(3 pi )/ 8$ and $\frac{π}{2}$ $( pi ) / 2$ . If one side of the triangle has a length of $9$ $9$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-11-17T18:01:33

$P e r i m e t e r = 23.5 + 21.7 + 9 = 54.2$ $Perimeter = 23.5 + 21.7 + 9 = 54.2$

Let $∠ C = \frac{π}{2}$ $angle C = pi/2$

Let $∠ B = \frac{3 π}{8}$ $angle B = (3pi)/8$

Then $∠ A = π - ∠ C - ∠ B = \frac{π}{8}$ $angle A = pi - angle C - angle B = pi/8$

Make the length 9 side be "a" so that it is opposite the smallest angle, A; this will give us the longest possible perimeter:

Let side $a = 9$ $a = 9$

Find the length of side b, using the Law of Sines:

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

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

$b = 9 \frac{sin (3 \frac{π}{8})}{sin (\frac{π}{8})} \approx 21.7$ $b = 9sin(3pi/8)/sin(pi/8) ~~ 21.7$

Because side c is opposite a right angle, we can use:

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

$c = \sqrt{9^{2} + {21.7}^{2}}$ $c = sqrt(9^2 + 21.7^2)$

$c \approx 23.5$ $c ~~ 23.5$

$P e r i m e t e r = 23.5 + 21.7 + 9 = 54.2$ $Perimeter = 23.5 + 21.7 + 9 = 54.2$

Two corners of a triangle have angles of 3π8 (3 pi )/ 8 and π2 ( pi ) / 2 . If one side of the triangle has a length of 9 9 , what is the longest possible perimeter of the triangle?