In $△ A B C$ right angle is at $B . A B = 3 c m A C = 6 c m$ determine $∠ B A C$ and $∠ A C B$ .

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Solution

$sin ∠ B C A = \frac{A B}{A C} = \frac{3}{6} = \frac{1}{2}$
$sin ∠ B C A = sin 30 °$
$∠ B C A = 30 °$
$∠ A C B = 30 °$
$∴ ∠ A B C + ∠ B A C + ∠ A C B = 180 °$ [sum of all angles in a triangle in $180 °$ ]
$∠ B A C + 90 ° + 30 ° = 180 °$
$∠ B A C = 60 °$

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