In $△ A B C, ∠ B = 90^{o}, A B = 3 c m, A C = 6 c m, ∠ A =$

30^{o}

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45^{o}

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60^{o}

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None of these

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Solution

The correct option is B $60^{o}$
$sin C = \frac{A B}{A C} = \frac{3}{6} = \frac{1}{2}$

But $sin 30^{o} = \frac{1}{2}$ , so, $∠ C = 30^{o}$

$∠ A = {(90 - ∠ C)}^{o} = {(90 - 30)}^{o} = 60^{o}$ .

So $60^{o}$ is correct.

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Similar questions

A B C

∠ B = 90^{o}

A B = 3 c m

A C = 6 c m

∠ C

∠ A

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

\frac{{tan}^{2} 60^{o} + 4 {cos}^{2} 45^{o} + 3 {sec}^{2} 30^{o} + 5 {cos}^{2} 90^{o}}{c o sec 30^{o} + sec 60^{o} - {cot}^{2} 30^{o}}

4 ({sin}^{4} 30^{o} + {cos}^{4} 60^{o}) - 3 ({cos}^{4} 45^{o} - {sin}^{2} 90^{o})

△ A B C, ∠ B = 90^{o}

A B = (x - 3) c m, B C = (x + 4) c m

A C = (x + 6) c m

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