Question

Va rog să mă ajutați ​

Answer 1

Explicație pas cu pas:

<B = 90°

$BC = 12 \sqrt{3}$

$\sin(C) = \frac{ \sqrt{3} }{2}$

folosim formula:

$\sin^{2} ( \alpha ) + \cos^{2} ( \alpha ) = 1$

$= > \cos^{2} (C) = 1 - \sin^{2} (C) = 1 - \frac{3}{4} = \frac{1}{4} \\ \cos(C) = \sqrt{ \frac{1}{4} }$

$= > \cos(C) = \frac{1}{2}$

$\cos(C) = \frac{BC}{AC} = > \frac{BC}{AC} = \frac{1}{2} \\ = > AC = 2 \times 12 \sqrt{3} = 24 \sqrt{3}$

$= > AC = 24 \sqrt{3}$

$AB^{2} = AC^{2} - BC^{2} = {(24 \sqrt{3} )}^{2} - {(12 \sqrt{3})}^{2} = 1296 = > AB = \sqrt{1296}$

$= > AB = 36$

$\tan(C) = \frac{AB}{BC} = \frac{36}{12 \sqrt{3}} = \sqrt{3}$

$= > \tan(C) = \sqrt{3}$

$\cot(C) = \frac{BC}{AB} = \frac{12 \sqrt{3} }{36} = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3}$

$= > \cot(C) = \frac{ \sqrt{3} }{3}$