In the adjoining figure $∠ A B C = ?$

45^{o}

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30^{o}

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60^{o}

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50^{o}

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Solution

The correct option is B $60^{o}$
Since ABC is right angled triangle.
$\Rightarrow \frac{A C}{A B} = tan (∠ A B C)$
$\Rightarrow \frac{100 \sqrt{3}}{100} = tan (∠ A B C)$
$\Rightarrow tan (∠ A B C) = \sqrt{3}$
$\Rightarrow ∠ A B C = 60^{0}$
Option C is correct.

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Similar questions

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

∠ C O A - ∠ B O C = 50^{o}

\frac{1 + {tan}^{2} 30^{o}}{1 - {tan}^{2} 30^{o}} + {csc}^{2} 60^{0} - {cos}^{2} 45^{o} + {sin}^{2} 45^{o} + \frac{1 + {cot}^{2} 60^{o}}{1 - {cot}^{2} 60^{o}}

= 5.5

∠

= 50^{o}

∠

= 60^{o}

∠

= 30^{o}

\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}}

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