A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T08:02:57

Largest possible area of the triangle is 14.6556

Explanation:

Given are the two angles $\frac{3 π}{8}$ $(3pi)/8$ and $\frac{π}{6}$ $pi/6$ and the length 1

The remaining angle:

$= π - (\frac{3 π}{8}) + \frac{π}{6}) = \frac{11 π}{24}$ $= pi - ((3pi)/8) + pi/6) = (11pi)/24$

I am assuming that length AB (12) 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 $= \frac{4^{2} \cdot sin (\frac{11 π}{24}) \cdot sin (\frac{3 π}{8})}{2 \cdot sin (\frac{π}{6})}$ $=( 4^2*sin((11pi)/24)*sin((3pi)/8))/(2*sin(pi/6))$

Area $= 14.6556$ $=14.6556$

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?

1 Answer

Explanation:

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{3 π}{8}$ $(3 pi )/ 8$ . If one side of the triangle has a length of $4$ $4$ , what is the largest possible area of the triangle?