The problem is as follows:
In figure 1. there is a circle as shown. The radius is equal to 10 inches and its center is labeled with the letter O. If $\measuredangle PC=30^{\circ}$. $\textrm{Find AB+BC}$.
The existing alternatives in my book are:
- $3\left( \sqrt{2}+\sqrt{6}\right)$
- $4\left( \sqrt{6}-\sqrt{2}\right)$
- $5\left( \sqrt{3}-\sqrt{2}\right)$
- $5\left( \sqrt{3}+\sqrt{2}\right)$
- $5\left( \sqrt{2}+\sqrt{6}\right)$
After analyzing the drawing the figure from below shows all all the relationships which I could found and it is summarized as follows:
The triangle $\textrm{COP}$ is isosceles since it shares the same side from the radius of the circle and since $\measuredangle PC=30^{\circ}$, then all is left to do is to apply the identity which it says that the sum of inner angles in a triangle must equate to $180^{\circ}$.
$$2x+30^{\circ}=180^{\circ}$$ $$x=\frac{150^{\circ}}{2}=75^{\circ}$$
Since $\measuredangle OCP = \measuredangle OPC$, its supplementary angle would become:
$$180^{\circ}-75^{\circ}=105^{\circ}$$
Since it is given from the problem:
$$\measuredangle COA = 90^{\circ}$$
therefore its complementary angle with $\measuredangle COP = 30^{\circ}$ would become into:
$$\measuredangle POA = 60^{\circ}$$
Since $PO = OA$ this would also make another isosceles triangle and by recurring to the previous identity:
$$2x+60^{\circ}=180^{\circ}$$ $$x=\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ}$$
Therefore the triangle POA is an equilateral one so,
$$\textrm{PA=10 inches}$$
As $\measuredangle OPB = 105 ^{\circ}$ and $\measuredangle OPA = 60^{\circ}$ then its difference is: $\measuredangle APB = 45^{\circ}$.
From this its easy to note that $\measuredangle PAB = 45^{\circ}$.
Since the vertex $\textrm{B}$ of the triangle $\textrm{ABP}$ is $\measuredangle = 90 ^{\circ}$. I did identified a special right triangle with the form $45^{\circ}-45^{\circ}-90^{\circ}$ or $\textrm{k, k,}\,k\sqrt{2}$.
By equating the newly found side $\textrm{PA = 10 inches}$ to $k\sqrt{2}$ this is transformed into:
$$k\sqrt{2} = 10$$
$$k = \frac{10}{\sqrt{2}}$$
From this is established that:
$$AB = \frac{10}{\sqrt{2}}$$
Since we have $\textrm{AB}$ we also know $\textrm{PB}$ as $AB = PB = \frac{10}{\sqrt{2}}$
Therefore all that is left to do is to find $\textrm{CP}$ as $CP+PB = BC$
To find $CP$ I used cosines law as follows:
$$a^{2}=b^{2}+c^{2}-2bc\,\cos A$$
Being a, b and c the sides of a triangle ABC and A the opposing angle from the side taken as a reference in the left side of the equation.
In this case
$$(CP)^{2}= 10^{2}+10^{2}-2(10)(10)\cos30^{\circ}$$ $$(CP)^{2}= 10^{2} \left(1+1-2\left(\frac{\sqrt{3}}{2}\right)\right)$$ $$CP = 10 \sqrt{ 2-\sqrt{3}}$$
Therefore $CP = 10 \sqrt{ 2-\sqrt{3}}$ and we have all the parts so the rest is just adding them up.
$$CP+PB= BC = 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$
$$AB= \frac{10}{\sqrt{2}}$$
$$AB + BC = \frac{10}{\sqrt{2}} + 10 \sqrt{ 2-\sqrt{3}} + \frac{10}{\sqrt{2}}$$
And that's how far I went, but from then on I don't know if what I did was correct or did I missed something? as my answer doesn't appear within the alternatives.
The best I could come up with by simplifying was:
$$\frac{10\sqrt{2}}{2}+10\sqrt{2-\sqrt{3}}+\frac{10\sqrt{2}}{2}$$
$$10\sqrt{2}+10\sqrt{2-\sqrt{3}}$$
$$10\left(\sqrt{2}+\sqrt{2-\sqrt{3}}\right)$$
and, that's it. But it doesn't seem to be in the choices given. Can somebody help me to find if did I do something wrong?. If a drawing is necessary please include one as I'm not savvy enough to notice these things easily.
Assuming your work is correct so far, I think we may be able to simplify further.
$$\sqrt{2-\sqrt3}=\sqrt{\frac{4-2\sqrt3}{2}}=\frac{\sqrt{\sqrt3^2-2(1)\sqrt3+1^2}}{\sqrt2}=\frac{\sqrt{(\sqrt3-1)^2}}{\sqrt2}=\frac{\sqrt3-1}{\sqrt2}$$.
Now let's see if that helps.
$$10\left(\sqrt2+\sqrt{2-\sqrt3}\right)=5\sqrt2\left(2+\sqrt3-1\right)=5(\sqrt2+\sqrt6)$$
Again, assuming everything you've done is correct, the last answer is the solution.