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Ant kubo ABCDA₁B₁C₁D₁ kraštinės CC₁ atidėtas taškas E taip, kad atkarpa EC₁ yra 2 kartus ilgesnė už atkarpą EC.

Raskite kampą tarp tiesių, einančių per BE ir AC₁.

Sprendimas:

Tarkime, kubo kraštinė lygi 1.

$AC_{1}^{2}$ = $1^{2}+1^{2}+1^{2}$ = $3$

Pažymime tašką F analogišką taškui E, t.y. FD = $\frac{1}{3}\cdot DD_{1}$ .

Ieškomas kampas lygus ∠FAC₁, nes, AF || BE

Pagal Pitagoro teoremą

$AF^{2}$ = $1^{2}+(\frac{1}{3})^{2}$ = $1+\frac{1}{9}$ = $\frac{10}{9}$

$FC_{1}^{2}$ = $1^{2}+(\frac{2}{3})^{2}$ = $1+\frac{4}{9}$ = $\frac{13}{9}$

Pagal kosinusų teoremą

$FC_{1}^{2}$ = $AF^{2}+AC_{1}^{2}-2\cdot AF\cdot AC_{1}\cdot cos(A)$

13

/ 9

=

10

/ 9

+3- 2* saknis(

10

/ 9

)* saknis(

3)* cos(A)

13

/ 9

=

10

/ 9

+3- 2* saknis(

10

/ 9

)* saknis(

3)* cos(A) $\frac{13}{9}$ = $\frac{10}{9}+3-2\cdot \sqrt {\frac{10}{9}}\cdot \sqrt {3}\cdot cos(A)$

2* saknis(

10

/ 9

)* saknis(

3)* cos(A)+

13

/ 9

=

10

/ 9

+3 $2\cdot \sqrt {\frac{10}{9}}\cdot \sqrt {3}\cdot cos(A)+\frac{13}{9}$ = $\frac{10}{9}+3$

2* saknis(

10

/ 9

)* saknis(

3)* cos(A) =

10

/ 9

-

13

/ 9

+3 $2\cdot \sqrt {\frac{10}{9}}\cdot \sqrt {3}\cdot cos(A)$ = $\frac{10}{9}-\frac{13}{9}+3$

2* saknis(

30)

/ 3

* cos(A) =

10

/ 9

-

13

/ 9

+3 $\frac{2\cdot \sqrt {30}}{3}\cdot cos(A)$ = $\frac{10}{9}-\frac{13}{9}+3$

2* saknis(

30)

/ 3

* cos(A) = -

1

/ 3

+3 $\frac{2\cdot \sqrt {30}}{3}\cdot cos(A)$ = $-\frac{1}{3}+3$

2* saknis(

30)

/ 3

* cos(A) =

8

/ 3

$\frac{2\cdot \sqrt {30}}{3}\cdot cos(A)$ = $\frac{8}{3}$

saknis(

30)

/ 3

* cos(A) =

8

/ 3/ 2

$\frac{\sqrt {30}}{3}\cdot cos(A)$ = $\frac{8}{3\cdot 2}$

saknis(

30)

/ 3

* cos(A) =

8

/ 6

$\frac{\sqrt {30}}{3}\cdot cos(A)$ = $\frac{8}{6}$

cos(A)* saknis(

30)

/ 3

=

8

/ 6

$\frac{cos(A)\cdot \sqrt {30}}{3}$ = $\frac{8}{6}$

cos(A)* saknis(

30) =

8* 3

/ 6

$cos(A)\cdot \sqrt {30}$ = $\frac{8\cdot 3}{6}$

cos(A) =

8* 3

/ 6/ saknis(

30)

$cos(A)$ = $\frac{8\cdot 3}{6\cdot \sqrt {30}}$

cos(A) =

4

/ saknis(

30)

$cos(A)$ = $\frac{4}{\sqrt {30}}$

cos(A) =

4* saknis(

30)

/ saknis(

30)/ saknis(

30)

$cos(A)$ = $\frac{4\cdot \sqrt {30}}{\sqrt {30}\cdot \sqrt {30}}$

cos(A) =

4* saknis(

30)

/ 30

$cos(A)$ = $\frac{4\cdot \sqrt {30}}{30}$

cos(A) =

2* saknis(

30)

/ 15

$cos(A)$ = $\frac{2\cdot \sqrt {30}}{15}$

arccos(cos(A)) = arccos(

2* saknis(

30)

/ 15

) $arccos(cos(A))$ = $arccos(\frac{2\cdot \sqrt {30}}{15})$

A = arccos(

2* saknis(

30)

/ 15

) $A$ = $arccos(\frac{2\cdot \sqrt {30}}{15})$

$\frac{13}{9}$ = $\frac{10}{9}+3-2\cdot \sqrt {\frac{10}{9}}\cdot \sqrt {3}\cdot cos(A)$

$2\cdot \sqrt {\frac{10}{9}}\cdot \sqrt {3}\cdot cos(A)$ = $\frac{10}{9}-\frac{13}{9}+3$

$\frac{2\cdot \sqrt {30}}{3}\cdot cos(A)$ = $\frac{8}{3}$

$\frac{cos(A)\cdot \sqrt {30}}{3}$ = $\frac{8}{6}$

$cos(A)$ = $\frac{4}{\sqrt {30}}$

$cos(A)$ = $\frac{4\cdot \sqrt {30}}{30}$

$cos(A)$ = $\frac{2\cdot \sqrt {30}}{15}$

$A$ = $arccos(\frac{2\cdot \sqrt {30}}{15})$

Atsakymas: $arccos(\frac{2\cdot \sqrt {30}}{15})$

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