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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{3}$ $(pi)/3$ . If side C has a length of $18$ $18$ and the angle between sides B and C is $\frac{π}{8}$ $( pi)/8$ , what are the lengths of sides A and B?

1 Answer

Oct 13, 2016

$| A | \approx 7.95$ $abs(A)~~7.95$
$| B | \approx 20.61$ $abs(B)~~20.61$

Explanation:

Given:
$XXX ∠ A : B = \frac{π}{3}$ $color(white)("XXX")/_A:B = pi/3$
$XXX ∠ B : C = \frac{π}{8}$ $color(white)("XXX")/_B:C=pi/8$
$XXX | C | = 18$ $color(white)("XXX")abs(C)=18$

$∠ A : B = \frac{π}{3}$ $/_A:B=pi/3$ and $∠ B : C = \frac{π}{8} X$ $/_B:C=pi/8color(white)("X")$
$XXX \to X ∠ C : A = π - (\frac{π}{3} + \frac{π}{8}) = \frac{13 π}{24}$ $color(white)("XXX")rarrcolor(white)("X")/_C:A = pi-(pi/3+pi/8) = (13pi)/24$

By the Law of Sines
$XXX \frac{| A |}{sin (∠ B : C)} = \frac{| B |}{sin (∠ C : A)} = \frac{| C |}{sin (∠ A : B)}$ $color(white)("XXX")abs(A)/sin(/_B:C) = abs(B)/sin(/_C:A)=abs(C)/sin(/_A:B)$

$\frac{| C |}{sin (∠ A : B)} = \frac{18}{sin (\frac{π}{3})} = 20.78461$ $abs(C)/sin(/_A:B) = 18/sin(pi/3) = 20.78461$

$| A | = 20.78461 \times sin (∠ B : C) = 20.78461 \times sin (\frac{π}{8}) = 7.953926$ $abs(A)=20.78461 xx sin(/_B:C) = 20.78461 xx sin(pi/8) =7.953926$

$| B | = 20.78461 \times sin (∠ C : A) = 20.78461 \times ξ n (\frac{13 π}{24}) = 20.60679$ $abs(B)=20.78461xxsin(/_C:A)=20.78461xxxin((13pi)/24) =20.60679$

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