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In Figure 1, we have three randomly drawn vectors $\\vec{A}$, $\\vec{B}$ and $\\vec{C}$ with magnitudes $A$, $B$ and $C$ and angles with the horizontal direction $a$, $b$ and $c$ respectively.

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If $A=\\var{A}\\,\\mathrm{cm}$, $B=\\var{B}\\,\\mathrm{cm}$ and $C=\\var{C}\\,\\mathrm{cm}$

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and $a=\\var{aa[i]}\\,\\mathrm{^o}$, $b=\\var{bb[j]}\\,\\mathrm{^o}$ and $c=\\var{cc[k]}\\,\\mathrm{^o}$

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fill in the gaps below.

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Use the following table:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
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angle($^o$)

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030456090120180
sin0$\\frac{1}{2}$$\\simplify{sqr(2)/2}$$\\simplify{sqr(3)/2}$1$\\simplify{sqr(3)/2}$0
cos1$\\simplify{sqr(3)/2}$$\\simplify{sqr(2)/2}$$\\frac{1}{2}$0-$\\frac{1}{2}$-1
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and round your answers to 2 decimal places.

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The sine values of aa.

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The cosine values of aa.

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Resolving $\\vec{A}$, $\\vec{B}$ and $\\vec{C}$ to their scalar components according to Fig. 2, we get

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i) $A_x$= [[0]] $\\,\\mathrm{cm}$     ii) $B_x$= [[2]] $\\,\\mathrm{cm}$     iii) $C_x$= [[4]] $\\,\\mathrm{cm}$

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   $A_y$= [[1]] $\\,\\mathrm{cm}$         $B_y$= [[3]] $\\,\\mathrm{cm}$          $C_y$= [[5]] $\\,\\mathrm{cm}$

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(0.5 marks each)

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Remember that for an angle $\\theta$ in a right triangle, by definition, $cos\\theta=\\frac{Adjacent Side}{Hypotenuse}$ while $sin\\theta=\\frac{Opposite Side}{Hypotenuse}$.

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If $\\vec{R}=\\var{coeff1}\\vec{A}-\\var{coeff2}\\vec{B}+\\var{coeff3}\\vec{C}$

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then in its vector component form, we will have:

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$\\vec{R}=$ [[0]]$\\hat{i}+$[[1]]$\\hat{j}$  (1.5 marks each)

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Use your answers from part (a) of this question.

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A vector $\\vec{A}$ in its vector component form is written as $A_x\\hat{i}+A_y\\hat{j}$.

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