A triangle has two corners with angles of $\frac{π}{12}$ $pi / 12$ and $\frac{7 π}{8}$ $(7 pi )/ 8$ . If one side of the triangle has a length of $6$ $6$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-05T12:27:00

Largest possible area of the triangle = 8.0031

Given are the two angles $\frac{7 π}{8}$ $(7pi)/8$ or ${157.5}^{\circ}$ $157.5^@$ and $\frac{π}{12}$ $pi/12$ or $15^{\circ}$ $15^@$ and the length 6

The remaining angle:

$180^{\circ} - ({157.5}^{\circ} + 15^{\circ}) = {17.5}^{\circ}$ $180^@-(157.5^@+15^@)=17.5^@$

I am assuming that length AB (6) 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 $= \frac{6^{2} \cdot sin (17.5) \cdot sin (157.5)}{2 \cdot sin (15)}$ $=( 6^2*sin(17.5)*sin(157.5))/(2*sin(15))$

Area $= 8.0031$ $=8.0031$

A triangle has two corners with angles of π12 pi / 12 and 7π8 (7 pi )/ 8 . If one side of the triangle has a length of 66 , what is the largest possible area of the triangle?