задача 24

[Сийка]

WIN_20190826_13_33_43_Pro (2).jpg (78.05 KiB) Прегледано 444 пъти

[Davids]

$tg\beta = \frac{1}{3} \Rightarrow cos\beta = 3sin\beta \Rightarrow sin^2\beta + 9sin^2\beta = 1 \Rightarrow sin\beta = \frac{1}{\sqrt{10}} \Rightarrow cos\beta = \frac{3}{\sqrt{10}}$
$\Rightarrow sin2\beta = 2sin\beta cos\beta = \frac{3}{5}$
$\Rightarrow cos2\beta = \frac{4}{5}$

Аналогично намираме за другия ъгъл:
$sin\alpha = \frac{1}{5\sqrt{2}}$
$cos\alpha = \frac{7}{5\sqrt{2}}$

Тогава $sin(\alpha + 2\beta) = sin\alpha cos2\beta + cos\alpha sin2\beta = \frac{1}{5\sqrt{2}}.\frac{4}{5} + \frac{7}{5\sqrt{2}}.\frac{3}{5} = \frac{1}{\sqrt{2}}$

Следователно стойността на сборния ъгъл е или 45°, или 135°. ^-^

[S.B.]

[tex]tg\alpha = \frac{1}{7},tg\beta =\frac{1}{3} ,\alpha ,\beta \in (0 ; \frac{\pi}{2})[/tex] търси се мярката на $ (\alpha + 2\beta)$
[tex]tg2\beta = \displaystyle\frac{2tg\beta}{1 - tg^{2}\beta} =\displaystyle \frac{2.\displaystyle\frac{1}{3}}{1 - \displaystyle\frac{1}{9}} = \frac{3}{4}[/tex]
$tg(\alpha + 2\beta) =\displaystyle \frac{tg\alpha + tg2\beta}{1 - tg\alpha.tg2\beta} = \displaystyle \frac{\displaystyle\frac{1}{7} + \displaystyle\frac{3}{4}}{1 - \displaystyle\frac{1}{7}.\displaystyle\frac{3}{4}} = 1$

$tg(\alpha + 2\beta) = 1 \Rightarrow \alpha + 2\beta = \frac{\pi}{4}$ или $ \frac{5 \pi}{4}$