Question

решить задачу по геометрии​

Answer 1

AB = 5BC

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Пусть BC = x, тогда AB = 5x

По теореме Пифагора :

$AC=\sqrt{BA^{2} -CB^{2} } =\sqrt{25x^{2} -x^{2} }=\sqrt{24x^{2} } =2x\sqrt{6}$

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1.

$sin$

2.

$cos < B = \frac{BC}{BA} = \frac{x}{5x} =\frac{1}{5}$

3.

$tg$

4.

$ctg < B = \frac{BC}{CA} = \frac{x}{2x\sqrt{6} } =\frac{1*\sqrt{6} }{2\sqrt{6}*\sqrt{6} }= \frac{\sqrt{6} }{12}$

Answer 2

AB=5BC

AC²=AB²-BC²

AC²=25BC²-BC²

AC²=24BC²

AC=2√6 BC

sin B=AC/AB

sin B=2√6BC/5BC=2√6/5

cos B=BC/AB

cos B=BC/5BC=1/5

tg B=AC/BC=2√6BC/BC=2√6

ctgB=BC/AC=BC/2√6BC=1/2√6=√6/12