How do you find the area of $△ A B C$ $triangle ABC$ given b=14.6, c=12.8, $m ∠ A = 56$ $mangleA=56$ ?

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How do you find the area of $△ A B C$ $triangle ABC$ given b=14.6, c=12.8, $m ∠ A = 56$ $mangleA=56$ ?

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Answer 1 · 2017-07-29T07:25:10

$77.5$ $77.5$

Explanation:

$Area of non right-angled triangle = \frac{a \cdot b \cdot sin C}{2}$ $"Area of non right-angled triangle" = (a*b*sinC)/2$ , where:
$a = 14.6$ $a = 14.6$
$b = 12.8$ $b=12.8$
$C = 56$ $C=56$

$\frac{12.8 \cdot 14.6 \cdot sin (56^{\circ})}{2} = 77.4652708 \approx 77.5$ $(12.8*14.6*sin(56^circ))/2=77.4652708~~77.5$

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How do you find the area of $△ A B C$ $triangle ABC$ given b=14.6, c=12.8, $m ∠ A = 56$ $mangleA=56$ ?

1 Answer

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How do you find the area of triangle ABC△ABCtriangle ABC given b=14.6, c=12.8, mangleA=56m∠A=56mangleA=56?

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How do you find the area of $△ A B C$ $triangle ABC$ given b=14.6, c=12.8, $m ∠ A = 56$ $mangleA=56$ ?