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 $1$ $1$ , 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 $1$ $1$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T12:50:02

Largest possible area of the triangle is 0.916

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 (1) 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 $= \frac{1^{2} \cdot sin (\frac{11 π}{24}) \cdot sin (\frac{3 π}{8})}{2 \cdot sin (\frac{π}{6})}$ $=( 1^2*sin((11pi)/24)*sin((3pi)/8))/(2*sin(pi/6))$

Area $= 0.916$ $=0.916$

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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 $1$ $1$ , what is the largest possible area of the triangle?

1 Answer

Explanation:

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A triangle has two corners with angles of π6 pi / 6 and 3π8 (3 pi )/ 8 . If one side of the triangle has a length of 11 , 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 $1$ $1$ , what is the largest possible area of the triangle?