Alan

The frictional force must exactly balance the horizontal force supplied by the string.  The tension in the string angled at 40° to the vertical is the same as the weight of C (because the pulley is frictionless).  The horizontal component of this is just 1kg*9.8m/s²*sin(40°)

$${\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{40}}^\circ\right)} = {\mathtt{6.299\: \!318\: \!574\: \!932\: \!6}}$$

or approximately 6.3N

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See if this helps:

(1)  10° + 32° + θ + Φ = 180°

(2) sin(32°)/CD = sin(θ)/12

(3) sin(Φ)/CD = sin(10°)/5

Three equations in three unknowns should enable you to find the angles.

Then use the sin rule again to find the lengths of the other sides, AC and BC.

Then use Herons formula to find the area:

Area = √[s(s - AB)(s - AC)(s - BC)]  where s = (AB + BC + AC)/2

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Here's an alternative method:

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$$y^3+3y^2-121y-363\rightarrow(y+3)(y+11)(y-11)$$

So the solutions are y = -3;  y = -11;  y = 11

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There's nothing wrong with your logic Melody; there are 3 possible solutions:

$${{\mathtt{a}}}^{{\mathtt{3}}} = {\left(-{\mathtt{3}}\right)}^{\left(-{\mathtt{4}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\
{\mathtt{a}} = {\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\\
{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\right)\\
{\mathtt{a}} = {\mathtt{0.231\: \!120\: \!424\: \!783\: \!544\: \!9}}\\
\end{array} \right\}$$

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Should there be an x in there somewhere?

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Calculators often use the following method to calculate fractional powers:

If we have a = b^c they take logs to get ln(a) = c*ln(b).  Now raise both sides to the power e to get a = e^c*ln(b).

If b is negative they can complain!

In fact: (-3)^-4/3 = -0.11556 + 0.200156i

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Does this help:

Let 1 - 1/x² = -1 and solve for x to get the limiting values of x at the -1 level

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If Z is the number of unit normal standard deviates, then the probability that Z>2.04 is 0.02068

Sam wasn't very good at maths, 1/2 + 1/3 + 1/12 is only 11/12, not 1.

However, add a dummy ax, so there are 12 axes and apply Sam's fractions to them.  Ron gets 12/2 = 6 axes; John gets 12/3 = 4 axes; the youngest gets 12/12 = 1 axe.

6 + 4 + 1 = 11, so this uses all Sam's axes and you can keep your own dummy axe!

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