A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , 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 $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-10-21T03:28:41

Largest possible area = 126.5489

Explanation:

Three angles are $\frac{π}{6}, \frac{5 π}{8}, \frac{5 π}{24}$ $pi/6, (5pi)/8, (5pi)/24$
Least angle is $\frac{π}{6}$ $pi/6$ which will correspond to side of length 15 to get maximum area.
$\frac{C}{sin (∠ C)} = \frac{A}{sin (∠ A)} = \frac{B}{sin (∠ B)}$ $C/ sin (/_C ) = A / sin( /_A ) = B / sin ( /_B)$
$\frac{C}{sin (\frac{5 π}{8})} = \frac{A}{sin (\frac{5 π}{24})} = \frac{15}{sin (\frac{π}{6})}$ $C/ sin ((5pi)/8) = A / sin ((5pi) / 24)= 15 / sin (pi/6)$

$C = \frac{15 \cdot sin (\frac{5 π}{8})}{sin (\frac{π}{6})} = 27.7164$ $C= (15 * sin ((5pi)/8))/sin (pi/6) = 27.7164$

$A = \frac{5 \cdot sin (\frac{5 π}{24})}{sin (\frac{π}{6})} = 18.2628$ $A = (5 * sin ((5pi)/24))/sin (pi/6) = 18.2628$

Semi Perimeter of $Delta ABC s = (15 + 27.7154 + 18.2628) / 2 = 30.49$

$s - a = 30.49 - 27.7164 = 2.7732$
$s - b = 30.49 - 15 = 15.49$
$s - c = 30.49 - 18.2628 = 12.2272$

Area of $Delta ABC = sqrt(s(s-a)(s-b)(s-c))$

Area of $Delta ABC = sqrt (30.49 * 2.7732 * 15.49 * 12.2272) = **126.5489**$

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , 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 $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?