Two corners of a triangle have angles of $\frac{3 π}{4}$ $(3 pi ) / 4$ and $\frac{π}{6}$ $pi / 6$ . If one side of the triangle has a length of $5$ $5$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-08T13:50:01

Largest possible area of the triangle is 17.0753

Given are the two angles $\frac{3 π}{4}$ $(3pi)/4$ and $\frac{π}{6}$ $pi/6$ and the length 5

The remaining angle:

$= π - ((\frac{3 π}{4}) + \frac{π}{6}) = \frac{π}{12}$ $= pi - (((3pi)/4) + pi/6) = pi/12$

I am assuming that length AB (5) is opposite the smallest angle.

Using the ASA

Area $= \frac{c^{2} \cdot sin (A) \cdot sin (B)}{2 \cdot sin (C)}$ $=(c^2*sin(A)*sin(B))/(2*sin(C)$

Area $= \frac{5^{2} \cdot sin (\frac{π}{6}) \cdot sin (\frac{3 π}{4})}{2 \cdot sin (\frac{π}{12})}$ $=( 5^2*sin(pi/6)*sin((3pi)/4))/(2*sin(pi/12))$

Area $= 17.0753$ $=17.0753$

Two corners of a triangle have angles of (3 pi ) / 4 3π4(3 pi ) / 4 and pi / 6 π6 pi / 6 . If one side of the triangle has a length of 5 55 , what is the longest possible perimeter of the triangle?