Геометрия: ів, розв яжіть задачі в повній формі(з дано, знайти)

Для простоти введемо позначення:A = CB, b= AC, c = AB, α = ∠A, β = ∠B1. b=3, cosα = 1/42. a = 5, sinα = 2/33. b = 8, tgβ = 34. c = 12, cosβ = 4/55. b = 6, cosβ = 1/36. c = 8, tgβ = 6/7...

ів, розв'яжіть задачі в повній формі(з дано, знайти)

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Ответ:

warfacegamer01

12.10.2020 06:55

Для простоти введемо позначення:

A = CB, b= AC, c = AB, α = ∠A, β = ∠B

1. b=3, cosα = 1/4

$cos\alpha =\frac{b}{c} = c =\frac{b}{cos\alpha } \\\\c = 3:\frac{1}{4} = 3\cdot 4 = 12\:\: (cm)\\\\a= \sqrt{12^2-3^3} = \sqrt{144-9} = \sqrt{135} \approx 11,6 \:\:(cm)$

2. a = 5, sinα = 2/3

$sin\alpha = \frac{a}{c} = c=\frac{a}{sin\alpha} \\\\c=5:\frac{2}{3} =\frac{5\cdot 3}{2} = \frac{15}{2}=7,5 \:\: (cm) \\\\b=\sqrt{(\frac{15}{2})^2-5^2 } = \sqrt{\frac{225}{5}-25 } = \sqrt{\frac{125}{4}}=\frac{5\sqrt{5} }{2} \approx 5,6 \:\: (cm)$

3. b = 8, tgβ = 3

$tg\beta =\frac{b}{a} = a=\frac{b}{tg\beta } \\\\a=\frac{8}{3} \approx 2,7 \:\: (cm)\\\\c=\sqrt{8^2+(\frac{8}{3} )^2} = \sqrt{64+\frac{64}{9} } = \sqrt{\frac{640}{9} }=\frac{8\sqrt{10} }{3} \approx 8,4 \:\: (cm)$

4. c = 12, cosβ = 4/5

$cos\beta =\frac{a}{c} = a=c\cdot cos\beta \\\\a = 12\cdot \frac{4}{5} = \frac{48}{5} = 9,6 \:\: (cm)\\\\b=\sqrt{12^2-(\frac{48}{5})^2 }= \sqrt{144-\frac{2304}{25} }= \sqrt{\frac{3600-2304}{25} }=\sqrt{\frac{1296}{25} }={\frac{36}{5} = 7,2 \:\: (cm)$

5. b = 6, cosβ = 1/3

$sin^2\beta +cos^2\beta =1 = sin^2\beta = 1- cos^2\beta \\\\sin^2\beta = 1-(\frac{1}{3} )^2 = 1-\frac{1}{9}=\frac{8}{9} ;\:\: sin\beta = \sqrt{\frac{8}{9} } =\frac{\sqrt{8}}{3} \\\\sin\beta=\frac{b}{c} = c = \frac{b}{sin\beta } \\\\c = 6:\frac{\sqrt{8}}{3} = 6\cdot \frac{3}{\sqrt{8}} = \frac{18\sqrt{8} }{\sqrt{8}\cdot \sqrt{8}} = \frac{18\sqrt{8}}{8} = \frac{9\sqrt{8}}{4} \approx 6,4 \:\: (cm)\\\\$

$a = \sqrt{(\frac{9\sqrt{8}}{4} )^2-6^2}= \sqrt{ \frac{81\cdot 8}{16} -36}= \sqrt{ \frac{648-576}{16}}=\sqrt{ \frac{72}{16}}= \frac{\sqrt {4}\cdot \sqrt {18}}{\sqrt{16}}= \frac{2 \sqrt {18}}{4}= \frac{\sqrt {2}\cdot \sqrt {9}}{2} = \frac{3\sqrt {2}}{2} \approx 2,1 \:\: (cm)$

6. c = 8, tgβ = 6/7

$1+tg^2\beta = \frac{1}{cos^2\beta } = cos^2\beta = \frac{1}{1+tg^2\beta } \\\\cos^2\beta = \frac{1}{1+(\frac{6}{7} )^2 } = \frac{1}{1+\frac{36}{49} } = \frac{1}{\frac{85}{49} } = \frac{49}{85} \\ cos\beta = \sqrt{\frac{49}{85}} = \frac{7}{\sqrt{85}} \\\\cos\beta = \frac{a}{c} = a = c\cdot cos\beta \\\\a = 8\cdot \frac{7}{\sqrt{85}}= \frac{56}{\sqrt{85}} \\$

$b=\sqrt{(8^2-\left(\frac{56}{\sqrt{85}}\right)^2} =\sqrt{64-\frac{3136}{85}} =\sqrt{\frac{5440-3136}{85}} =\sqrt{\frac{2304}{85}} = \frac{48}{\sqrt{85}} \approx 5,2 \:\: (cm)$