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How do you solve the triangle if C=60 degrees, A= 12, c=15?

1 Answer

Oct 23, 2015

$∠ A ≅ {43.9}^{\circ}$ $/_A ~= 43.9^@$
$∠ B ≅ {76.1}^{\circ}$ $/_B ~= 76.1^@$
$b ≅ 16.8$ $b ~= 16.8$

Explanation:

The Law of Sines tells us
$XXX \frac{sin (A)}{a} = \frac{sin (B)}{b} = \frac{sin (C)}{c}$ $color(white)("XXX")sin(A)/a=sin(B)/b=sin(C)/c$

#sin(C)/c = sin(60^@)/15 = (sqrt(3)/2)/15 = sqrt(3)/30 ~= 0.057735

$sin (A) = \frac{sin (C)}{c} \cdot a = 0.057735 \cdot 12 = 0.69282$ $sin(A) = sin(C)/c*a = 0.057735*12 = 0.69282$
$∠ A = arcsin (sin (A)) = arcsin (0.69282) ≅ {43.85378}^{\circ}$ $/_A = arcsin(sin(A)) = arcsin(0.69282) ~= 43.85378^@$

$∠ B = 180^{\circ} - (∠ A + ∠ C) = 180^{\circ} - ({43.85378}^{\circ} + 60^{\circ}) = {76.14622}^{\circ}$ $/_B = 180^@ - (/_A + /_C) = 180^@ - (43.85378^@+60^@) = 76.14622^@$

$\frac{sin (B)}{b} = \frac{sin (C)}{c} = 0.057735$ $sin(B)/b =sin(C)/c=0.057735$
$\to b = \frac{sin (B)}{0.057735} = \frac{sin ({76.14622}^{\circ})}{0.057735} = 16.81655$ $rarr b = sin(B)/0.057735 = sin(76.14622^@)/0.057735 = 16.81655$

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