Alan

If you are talking about a right-angled triangle where x is one of the acute angles, then the other is given by tan^-1(16/34)

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{16}}}{{\mathtt{34}}}}\right)} = {\mathtt{25.201\: \!123\: \!645\: \!475^{\circ}}}$$

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Alternatively, you can do it by taking 90°-tan^-1(34/16)

$${\mathtt{90}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{34}}}{{\mathtt{16}}}}\right)} = {\mathtt{25.201\: \!123\: \!645\: \!475}}$$

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I've interpreted the words "missing angle" as the other acute angle in a right-angled triangle.  Melody is possibly more correct in interpreting it as just the value of x. (Though you should have 16 not 15 Melody!).

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Just substitute 27 for x in your expression. (27 = 2007 - 1980).

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$$\frac{3x(3x+21)}{(x+7)(9x^2)}=\frac{9x(x+7)}{(x+7)9x^2}=\frac{1}{x}$$

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Subtract 3x from both sides of the second equation:

y = 5 - 3x

Substitute this into the first equation:

x + 2(5 - 3x) = 5

or

x + 10 - 6x = 5

-5x = -5

x = 1

Substitute this back into the equation for y

y = 5 - 3*1

y = 2

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Here's one possible interpretation:

Of course this is just one possibility.  As far as the information provided in the question is concerned, T could be anywhere on a circle of radius 35cm centred on P!

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You will need a numerical method to find the values of y (in radians) that satisfy this equation.  The following graphs should enable you to get good guesses with which to start:

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In general (x^a)^b = x^(a*b) so

$$(x^\frac{1}{2})^a=x^1$$

$$x^{\frac{a}{2}}=x^1$$

We must have a/2 = 1, so a = 2

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This can be written as:

$$\frac{2}{\sqrt{x}}=3$$

Multiply both sides by the √x and divide both side by 3:

$$\frac{2}{3}=\sqrt{x}$$

Square both sides:

$$x=\frac{4}{9}$$

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$$2yy^{\frac{1}{3}}=0.0002$$

This can be written as:

$$2y^{1\frac{1}{3}}=0.0002$$

Divide both sides by 2

$$y^{1\frac{1}{3}}=0.0001$$

Take log₁₀ of both sides and use the fact that log(x^a) = a*log(x):

$$1\frac{1}{3}\log{y}=\log{10^{-4}}$$

as 0.0001 = 10^-4

Now log(10^-4) is just -4, so

$$1\frac{1}{3}\log{y}=-4$$

or:

$$\frac{4}{3}\log{y}=-4$$

Multiply both sides by 3/4:

$$\log{y}=-3$$

So: y = 10^-3 or y = 0.001

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