Alan

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You need to supply a diagram or explain in more detail how W, T¹ and T² relate to each other and to m and A.

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A half of one quarter is one eighth, so:

$$\frac{1}{3}+\frac{1}{8}\rightarrow \frac{8+3}{3\times 8}\rightarrow \frac{11}{24}$$ 

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This must be of the form a/60 + b/60 + c/60 = 47/60

But since it must also be of the form 1/p + 1/q + 1/r = 47/60 then a, b and c must be factors of 60 (so that each term on the LHS cancels to leave a 1 in the numerator).

$${factor}{\left({\mathtt{60}}\right)} = {{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{5}}$$

Clearly the 3 consecutive numbers are 3, 4 and 5, so their sum is 12.

Check

$${\frac{{\mathtt{1}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}} = {\frac{{\mathtt{47}}}{{\mathtt{60}}}} = {\mathtt{0.783\: \!333\: \!333\: \!333\: \!333\: \!3}}$$

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0.0254 is the number of metres/inch so turn 5' 2" into inches, namely 62".

Now multiply this by 0.0254 to get 0.0254metres/inch*62inches = 1.5748 metres

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$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left(-{\mathtt{0.939\: \!69}}\right)} = -{\mathtt{69.999\: \!560\: \!966\: \!236^{\circ}}}$$

Let's call it -70°.  Sine is negative in the third and fourth quadrants, so the angles are  360°-70° and 180°+70°

or 290° and 250°

Check:

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{290}}^\circ\right)} = -{\mathtt{0.939\: \!692\: \!620\: \!786}}$$ 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{250}}^\circ\right)} = -{\mathtt{0.939\: \!692\: \!620\: \!786}}$$

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Rather than writing all the other terms involving x², x³ etc. I've written O(x²), meaning terms of order x² and smaller ("smaller" because we are going to make x go to 0, so x⁴, x³ etc. are going to be smaller than x²). The important thing is that all these other terms will be multiples of xⁿ, where n>1, so that when they are divided by x they are still mutiples of powers of x and hence vanish in the limit as x tends to 0.

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Here's a derivation explicitly using the binomial expansion:

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The n'th term is given by a*r^n-1 so you want n such that 5*2^n-1 is just greater than 500.

The numbers are small enough to see that if n = 7, then 5*2⁶ = 5*64 is less than 500 and if n = 8 then 5*2⁷ = 5*128 is greater than 500 so the eighth term is the first one greater than 500 and is 5*128 = 640

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