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 ∠QTA=41∘ and ∠RUA=63∘, then find ∠QPR, in degrees.
3. Points A and B lie on a circle centered at O, and point P is outside the circle such that ¯AP and ¯BP are tangent to the circle. If ∠APO=26∘, then what is the measure of minor arc AB, in degrees?
Thank you!
+659 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. ADAB=BEBC
Subsitute in known values.
2. 317=7BC
Can you solve for BC?
+659 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 = Now you do the rest!
Shouldn't be that hard, considering that you read my solution.
+659 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 = Now you can do it!
