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1. A circle centered at $$A$$ with radius $$10$$ is externally tangent to a circle centered at $$B$$ with radius $$7.$$ A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line $$AB$$ at $$C$$ Find the length $$BC$$.

2. Lines $$PTQ$$ and $$PUR$$ are tangent to a circle, as shown below.

If $$\angle QTA = 41^\circ$$ and $$\angle RUA = 63^\circ,$$ then find $$\angle QPR,$$ in degrees.

3. Points $$A$$ and $$B$$ lie on a circle centered at $$O$$, and point $$P$$ is outside the circle such that $$\overline{AP}$$ and $$\overline{BP}$$ are tangent to the circle. If $$\angle APO = 26^{\circ}$$, then what is the measure of minor arc $$AB$$, in degrees?

Thank you!

Mar 29, 2020

#1
+626
+1

Question 1:

(I drew this)

AD = 10 - 7 = 3

Through a series of proofs, we know that ADB is similar to BEC by AA.

We set up a proportion.

1. $$\frac{AD}{AB}=\frac{BE}{BC}$$

Subsitute in known values.

2. $$\frac{3}{17}=\frac{7}{BC}$$

Can you solve for BC?

Mar 29, 2020
#2
+626
+2

Question 2:

Draw:

We draw TO, OU, and OA so that they meet at the center as radii.

Plan with what you know:

Since angle QTA = 41 and angle RUA = 63, we can use that information to find angle TOU.

TO, OA, and OU are all congruent as they are radii. Because of this, isosceles triangles TOA and UOA are formed.

Solving:

PUA = 180 - RUA = 117

PTA = 180 - QTA = 139

OUA = PUA - 90 = 27

OTA = PTA - 90 = 49

UOA = 180 - 2(OUA) = 126

TOA = 180 - 2(OTA) = 82

TOU = 360 - (UOA + TOA) = 152

QPR = 360 - 90 - 90 - TOU = $$\boxed{\text{Now you do the rest!}}$$

Shouldn't be that hard, considering that you read my solution.

Mar 29, 2020
#3
+626
+1

Question 3:

(I drew this)

Consider:

Angle APO = 26 and the right angles formed by tangent and radii lines

By the Two-Tangent theorem, BP = PA

By a serious of proofs, we know that triangles BPO and APO are congruent by SSS.

Solve:

Because of CPCTC, we know that BPO is congruent to APO and is therefore also 26 degrees.

We solve for BOP and AOP

BOP = 180 - 26 - 90 = 64

AOP = 180 - 26 - 90 = 64

BOA = 64 + 64

Minor arc AB = $$\boxed{\text{Now you can do it!}}$$

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Mar 29, 2020
#4
0