mjhjmtu

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 $measuredangle PC=30^circ$. $textrmFind AB+BC$.

The existing alternatives in my book are:

• $3left( sqrt2+sqrt6right)$


• $4left( sqrt6-sqrt2right)$


• $5left( sqrt3-sqrt2right)$


• $5left( sqrt3+sqrt2right)$


• $5left( sqrt2+sqrt6right)$

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 $textrmCOP$ 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 $180^circ$.

$$2x+30^circ=180^circ$$
$$x=frac150^circ2=75^circ$$

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:

$$2x+60^circ=180^circ$$
$$x=frac180^circ-60^circ2=60^circ$$

Therefore the triangle POA is an equilateral one so,

$$textrmPA=10 inches$$

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 $textrmB$ of the triangle $textrmABP$ is $measuredangle = 90 ^circ$. I did identified a special right triangle with the form $45^circ-45^circ-90^circ$ or $textrmk, k,,ksqrt2$.

By equating the newly found side $textrmPA = 10 inches$ to $ksqrt2$ this is transformed into:

$$ksqrt2 = 10$$

$$k = frac10sqrt2$$

From this is established that:

$$AB = frac10sqrt2$$

Since we have $textrmAB$ we also know $textrmPB$ as $AB = PB = frac10sqrt2$

Therefore all that is left to do is to find $textrmCP$ 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

$$(CP)^2= 10^2+10^2-2(10)(10)cos30^circ$$
$$(CP)^2= 10^2 left(1+1-2left(fracsqrt32right)right)$$
$$CP = 10 sqrt 2-sqrt3$$

Therefore $CP = 10 sqrt 2-sqrt3$ and we have all the parts so the rest is just adding them up.

$$CP+PB= BC = 10 sqrt 2-sqrt3 + frac10sqrt2$$

$$AB= frac10sqrt2$$

$$AB + BC = frac10sqrt2 + 10 sqrt 2-sqrt3 + frac10sqrt2$$

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:

$$frac10sqrt22+10sqrt2-sqrt3+frac10sqrt22$$

$$10sqrt2+10sqrt2-sqrt3$$

$$10left(sqrt2+sqrt2-sqrt3right)$$

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.

Notice that $$angle ACB=angle OCB-angle OCA=75^o-45^o=30^o,$$and $$AC=10sqrt2.$$Thus $$AB+BC=ACcdot(sin angle ACB+cos angle ACB)=5(sqrt2+sqrt6).$$

Assuming your work is correct so far, I think we may be able to simplify further.

$$sqrt2-sqrt3=sqrtfrac4-2sqrt32=fracsqrtsqrt3^2-2(1)sqrt3+1^2sqrt2=fracsqrt(sqrt3-1)^2sqrt2=fracsqrt3-1sqrt2$$.

Now let's see if that helps.

$$10left(sqrt2+sqrt2-sqrt3right)=5sqrt2left(2+sqrt3-1right)=5(sqrt2+sqrt6)$$

Again, assuming everything you've done is correct, the last answer is the solution.

I have checked your work and it is correct.

Your answer is correct and $$10left(sqrt2+sqrt2-sqrt3 right)=5left( sqrt2+sqrt6right)=19.31851653$$