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

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

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Answer 1 · 2018-04-19T07:46:01

$Largest possible Area of Triangle A_{t} = 87.43 sq units$ $color(maroon)("Largest possible Area of Triangle " A_t = 87.43 " sq units"$

Explanation:

$ˆ A = \frac{π}{4}, ˆ B = \frac{π}{6}, ˆ C = π - \frac{π}{4} - \frac{π}{6} = \frac{7 π}{12}$ $hat A = pi/4, hat B = pi/6, hat C = pi - pi/4 - pi/6 = (7pi)/12$

To get the largest area, side 8 should correspond to least angle $ˆ B$ $hat B$ and hence is side b.

Applying the Law of Sines,

$\frac{a}{sin A} = \frac{b}{sin B}$ $a / sin A = b / sin B$

$a = \frac{8 \cdot sin (\frac{π}{4})}{sin (\frac{π}{6})} = \frac{16}{\sqrt{2}}$ $a = (8 * sin (pi/4)) / sin (pi/6) = 16/sqrt2$

$Largest possible Area of Triangle A_{t} = (\frac{1}{2}) a b sin C$ $"Largest possible Area of Triangle " A_t = (1/2) a b sin C$

$A_{t} = (\frac{1}{2}) \cdot (\frac{16}{\sqrt{2}}) \cdot 8 \cdot sin (\frac{7 π}{12}) = 87.43 sq units$ $A_t = (1/2) * (16/sqrt2) * 8 * sin ((7pi)/12) = 87.43 " sq units"$

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

1 Answer

Explanation:

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