Two corners of a triangle have angles of $\frac{π}{3}$ $( pi )/ 3$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $7$ $7$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-10-15T23:02:01

The longest possible perimeter is 25.13

Compute remaining angle, C, is:

$C = π - \frac{π}{4} - \frac{π}{3}$ $C = pi - pi/4 - pi/3$

$C = \frac{12 π}{12} - \frac{3 π}{12} - \frac{4 π}{12}$ $C = (12pi)/12 - (3pi)/12 - (4pi)/12$

$C = \frac{5 π}{12}$ $C = (5pi)/12$

Let angle $A$ $A$ equal the smallest angle, $\frac{π}{4}$ $pi/4$ , then angle $B = \frac{π}{3}$ $B = (pi)/3$ .

Let 7 be the length of the side opposite angle A (we represent opposite sides with corresponding lowercase letters), $a = 7$ $a = 7$ .

When you do this, using the Law of Sines,

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

, to compute the lengths of sides b and c will give the sides that are the largest perimeter.

$b = 7 \frac{sin (\frac{π}{3})}{sin (\frac{π}{4})}$ $b = 7sin(pi/3)/sin(pi/4)$

$b \approx 8.57$ $b ~~ 8.57$

$c = 7 \frac{sin (\frac{5 π}{12})}{sin (\frac{π}{4})}$ $c = 7sin((5pi)/12)/sin(pi/4)$

$c \approx 9.56$ $c ~~ 9.56$

$p = 7 + 8.57 + 9.56$ $p = 7 + 8.57 + 9.56$

$p = 25.13$ $p = 25.13$

Two corners of a triangle have angles of π3 ( pi )/ 3 and π4 ( pi ) / 4 . If one side of the triangle has a length of 7 7 , what is the longest possible perimeter of the triangle?