Question

Срочно помогите
Геометрия

Answer 1

держи, вроде все верно

Answer 2

$1) \: \sin( \alpha ) = \frac{BC}{AB} = \frac{12}{15} = \frac{4}{5}$

$\cos( \alpha ) = \frac{AC}{AB} = \frac{9}{15} = \frac{3}{5}$

$tg( \alpha ) = \frac{BC}{AC} = \frac{12}{9} = 1 \frac{1}{3}$

$ctg( \alpha ) = \frac{AC}{BC} = \frac{9}{12} = \frac{3}{4}$

2) По теореме Пифагора:

с² = a² + b²

$c = \sqrt{a² + b²</p><p>} = \sqrt{ {8}^{2} + {15}^{2} } =$

$= \sqrt{289} = 17$

Ответ: 17 см.

3) sin² a + cos²a = 1

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

$= \sqrt{1 - { (\frac{5}{13}) }^{2} } = \sqrt{1 - \frac{25}{169} } =$

$= \sqrt{ \frac{144}{169} } = \frac{12}{13}$

$tg \alpha = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{12}{13} \div \frac{5}{13} = \frac{12}{13} \times \frac{13}{5} =$

$= 2 \frac{2}{5}$

$ctg \alpha = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = \frac{5}{12}$

$4) \: \cos {}^{2} \alpha + tg \alpha \times \cos {}^{2} \alpha = \\ = \cos {}^{2} \alpha+ \frac{ \sin {}^{2} \alpha }{ \cos {}^{2} \alpha } \times \cos {}^{2} \alpha = \\ = \cos {}^{2} \alpha+ \sin {}^{2} \alpha = 1$