Matematik

vektorer

31. januar 2021 af august543 - Niveau: B-niveau

Hej:)

Jeg ved ikke hvordan jeg skal gribe nogen af delopgaverne an i den vedhæftede fil. Er der muligvis nogen der kan forklare?

Tak på forhånd!


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Svar #1
31. januar 2021 af mathon


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Svar #2
31. januar 2021 af mathon

         \small \begin{array}{lllll} \textbf{a)}\\& \begin{array}{lllll}& \tan(90\degree -69\degree)=\frac{y_C}{13}\\\\& y_C=13\cdot \tan(21\degree)\\\\& C(13,13\cdot \tan(21\degree)) \end{array}\\\\\\ \textbf{b)}\\& \begin{array}{lllll} \left | EC \right |=\sqrt{13^2+\left ( 13\cdot \tan(21\degree) \right )^2}=13.9249\\ \textup{som er banetovets l\ae ngde.} \end{array}\\\\\\ \textbf{c)}\\& \begin{array}{lllll} \overrightarrow{e}=\begin{bmatrix} \cos(90\degree -(69\degree+13\degree))\\ \sin(90\degree -(69\degree+13\degree)) \end{pmatrix}=\begin{bmatrix} \cos(8\degree)\\ \sin(8\degree) \end{bmatrix} \end{array}\\\\\\ \textbf{d)}\\& \begin{array}{lllll} T_{CDE}=\frac{\left | CE \right |}{2}\cdot \frac{\sin(13\degree)\cdot \sin(16\degree)}{\sin(151\degree)}=0.8905 \end{array}\end{array}


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Svar #3
31. januar 2021 af mathon

detaljer for bergning i d)

        anvendt er formlen:
                                           \small \begin{array}{lllll} T_{ABC}=\frac{1}{2}\cdot a\cdot b\cdot \sin(C)=\frac{a}{2}\cdot a\cdot \frac{\sin(B)}{\sin(A)}\cdot \sin(C)=\frac{a^2}{2}\cdot \frac{\sin(B)\cdot \sin(C)}{\sin(A)} \end{array}
                                  


Svar #4
31. januar 2021 af august543

#2

         \small \begin{array}{lllll} \textbf{a)}\\& \begin{array}{lllll}& \tan(90\degree -69\degree)=\frac{y_C}{13}\\\\& y_C=13\cdot \tan(21\degree)\\\\& C(13,13\cdot \tan(21\degree)) \end{array}\\\\\\ \textbf{b)}\\& \begin{array}{lllll} \left | EC \right |=\sqrt{13^2+\left ( 13\cdot \tan(21\degree) \right )^2}=13.9249\\ \textup{som er banetovets l\ae ngde.} \end{array}\\\\\\ \textbf{c)}\\& \begin{array}{lllll} \overrightarrow{e}=\begin{bmatrix} \cos(90\degree -(69\degree+13\degree))\\ \sin(90\degree -(69\degree+13\degree)) \end{pmatrix}=\begin{bmatrix} \cos(8\degree)\\ \sin(8\degree) \end{bmatrix} \end{array}\\\\\\ \textbf{d)}\\& \begin{array}{lllll} T_{CDE}=\frac{\left | CE \right |}{2}\cdot \frac{\sin(13\degree)\cdot \sin(16\degree)}{\sin(151\degree)}=0.8905 \end{array}\end{array}

kan du måske forklare hvorfor man kan udregne det som du har gjort? Forstår det nemlig ikke, men tusind tak fordi du hjælper!:)


Svar #5
31. januar 2021 af august543

#2

         \small \begin{array}{lllll} \textbf{a)}\\& \begin{array}{lllll}& \tan(90\degree -69\degree)=\frac{y_C}{13}\\\\& y_C=13\cdot \tan(21\degree)\\\\& C(13,13\cdot \tan(21\degree)) \end{array}\\\\\\ \textbf{b)}\\& \begin{array}{lllll} \left | EC \right |=\sqrt{13^2+\left ( 13\cdot \tan(21\degree) \right )^2}=13.9249\\ \textup{som er banetovets l\ae ngde.} \end{array}\\\\\\ \textbf{c)}\\& \begin{array}{lllll} \overrightarrow{e}=\begin{bmatrix} \cos(90\degree -(69\degree+13\degree))\\ \sin(90\degree -(69\degree+13\degree)) \end{pmatrix}=\begin{bmatrix} \cos(8\degree)\\ \sin(8\degree) \end{bmatrix} \end{array}\\\\\\ \textbf{d)}\\& \begin{array}{lllll} T_{CDE}=\frac{\left | CE \right |}{2}\cdot \frac{\sin(13\degree)\cdot \sin(16\degree)}{\sin(151\degree)}=0.8905 \end{array}\end{array}

hvorfor trækker man 69 fra 90 i opgave a?


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Svar #6
31. januar 2021 af ringstedLC

a) Forlæng BC ned på x-aksen og kald skæringen for F.

Trekant CEF er retvinklet og vinkel CEF = 90º - 69º.

b) 

\begin{align*} \left |\overrightarrow{CE} \right | &= \left |\overrightarrow{EC} \right | =\left |\overrightarrow{OC} \right |\;,\;\overrightarrow{OC}=\,\text{stedvektoren for}\;C \\ \overrightarrow{EC} &= \binom{13}{13\cdot \tan(21^{\circ})}=\binom{x_C}{y_C} \\ \left |\overrightarrow{EC} \right | &= \sqrt{13^2+\bigl(13\cdot \tan(21^{\circ}\bigr)^2} \end{align*}

eller med ét ord: Pythagoras!

c) Her bruges:

\begin{align*} \overrightarrow{e}_{\overrightarrow{ED}} &= \text{enhedsvektoren for}\;\overrightarrow{ED} \\ &= \frac{1}{\left |\overrightarrow{ED}\right |}\cdot \overrightarrow{ED} =\binom{\cos(v)}{\sin(v)}\;,\;v=\angle DEF=90^{\circ}-(69^{\circ}+13^{\circ})=8^{\circ} \end{align*}

d) Længden af ED findes med sinusrelationerne:

\begin{align*} \angle C &= 180^{\circ}-(\angle E+\angle D) \\ &= 180^{\circ}-(13^{\circ}+151^{\circ})=16^{\circ} \\ \frac{\sin\bigl(\angle D\bigr)}{\left |EC \right |} &= \frac{\sin\bigl(\angle C\bigr)}{\left |ED \right |} \Rightarrow \left |ED \right |=\frac{\sin\bigl(\angle C\bigr)\cdot \left |EC \right |}{\sin\bigl(\angle D\bigr)}=\;? \\ \overrightarrow{e}_{\overrightarrow{ED}} &= \binom{\cos(8^{\circ})}{\sin(8^{\circ})} \Rightarrow \overrightarrow{ED}=\left |\overrightarrow{ED}\right |\cdot \binom{\cos(8^{\circ})}{\sin(8^{\circ})} =\binom{|\overrightarrow{ED}|\cdot \cos(8^{\circ})}{|\overrightarrow{ED}|\cdot \sin(8^{\circ})} \\ A_{\Delta CDE} &= \tfrac{1}{2}\cdot \left | \text{det}\Bigl(\overrightarrow{EC},\overrightarrow{ED}\Bigr) \right| \\ &= 12.4\; (m^2) \end{align*}


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Svar #7
31. januar 2021 af mathon

korrektion:

         \small \small \begin{array}{lllll} \textbf{a)}\\& \begin{array}{lllll}& \tan(90\degree -69\degree)=\frac{y_C}{13}\\\\& y_C=13\cdot \tan(21\degree)\\\\& C(13,13\cdot \tan(21\degree)) \end{array}\\\\\\ \textbf{b)}\\& \begin{array}{lllll} \left | EC \right |=\sqrt{13^2+\left ( 13\cdot \tan(21\degree) \right )^2}=13.9249\\ \textup{som er banetovets l\ae ngde.} \end{array}\\\\\\ \textbf{c)}\\& \begin{array}{lllll} \overrightarrow{e}=\begin{bmatrix} \cos(90\degree -(69\degree+13\degree))\\ \sin(90\degree -(69\degree+13\degree)) \end{pmatrix}=\begin{bmatrix} \cos(8\degree)\\ \sin(8\degree) \end{bmatrix} \end{array}\\\\\\ \textbf{d)}\\& \begin{array}{lllll} T_{CDE}=\frac{\left | CE \right |\mathbf{{\color{Red} ^2}}}{2}\cdot \frac{\sin(13\degree)\cdot \sin(16\degree)}{\sin(151\degree)}=12.4 \end{array}\end{array}


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