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 $8$ $8$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-07-19T06:43:33

The area is $= 2.75$ $=2.75$

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The angles ot the triangle are

$ˆ A = \frac{1}{6} π$ $hatA=1/6pi$

$ˆ B = \frac{3}{8} π$ $hatB=3/8pi$

$ˆ C = π - (\frac{1}{6} π + \frac{3}{8} π) = π - \frac{13}{24} π = \frac{11}{24} π$ $hatC=pi-(1/6pi+3/8pi)=pi-13/24pi=11/24pi$

The side of length is opposite the smallest angle in the triangle

So,

$a = 6$ $a=6$

We apply the sine rule to the triangle

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

$\frac{b}{sin (\frac{3}{8} π)} = \frac{6}{sin (\frac{1}{6} π)}$ $b/sin(3/8pi)=6/sin(1/6pi)$

$b = 6 \cdot \frac{sin (\frac{3}{8} π)}{sin (\frac{1}{6} π)} = 11.1$ $b=6*sin(3/8pi)/sin(1/6pi)=11.1$

The area of the triangle is

$a r e a = \frac{1}{2} | \in | ˆ C = \frac{1}{2} \cdot 6 \cdot 11.1 \cdot sin (\frac{11}{24} π)$ $area =1/2absin hatC=1/2*6*11.1*sin(11/24pi)$

$= 2.75$ $=2.75$

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 88 , what is the largest possible area of the triangle?