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Angles Inside and Outside Circles: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A vertical pair inside a circle

Points $A,B,C,D$ occur in that order around a circle. Chords $AC,BD$ cross at interior $P$ . Consecutive arcs $AB,CD$ measure $64^\circ,118^\circ$ . Find $m\angle APB$ and $m\angle CPD$ .

Schematic; use the stated givens.

Text description of this figure

Chords AC and BD cross at interior P; blue arc AB and dashed red arc CD face vertical angles. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  Locate the two arcs facing the vertical pair at P.
2. Hint 2
  Add their measures, then divide by two. Vertical angles are equal.
Answer

$\begin{gathered}m\angle APB=91^\circ,\\m\angle CPD=91^\circ.\end{gathered}$

Full solution

Angle APB uses consecutive arcs AB and CD.
Their half-sum is
$(64^\circ+118^\circ)/2=91^\circ.$

Angle CPD is vertical to APB, so it also measures $91^\circ$ .
The unspecified arcs BC and DA have a positive total of $178^\circ$ ; for instance, each can measure $89^\circ$ .

Answer

$\begin{gathered}m\angle APB=91^\circ,\\m\angle CPD=91^\circ.\end{gathered}$

Key idea

An inside angle uses the arcs facing its vertical pair.
Problem 2 Follow the external rays

Two secants from external $P$ have orders $P,A,B$ and $P,D,C$ . The circle order is $A,B,C,D$ . Far arc $BC$ excludes $A,D$ ; near arc $DA$ excludes $B,C$ . The far arc measures $176^\circ$ and the near arc measures $48^\circ$ . Point $Y$ lies on the first secant with order $Y,P,A,B$ .

Find $m\angle APD$ .

Find $m\angle YPD$ .

Schematic; use the stated givens.

Text description of this figure

External P has secant ray orders P,A,B and P,D,C. Blue far arc BC excludes A,D; dashed red near arc DA excludes B,C. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  First measure the angle between the two rays that point toward the circle.
2. Hint 2
  Ray PY reverses ray PA, so the second requested angle is supplementary to the first.
Answer

$\begin{gathered}m\angle APD=64^\circ,\\m\angle YPD=116^\circ.\end{gathered}$

Full solution

The far endpoints are B,C and the near endpoints are A,D.
Hence
$m\angle APD=(176^\circ-48^\circ)/2.$

The angle APD is $64^\circ$ .
Since PA and PY are opposite,
$m\angle YPD=180^\circ-64^\circ.$

The second angle is $116^\circ$ .
The other two consecutive arcs can each be $68^\circ$ , completing a positive $360^\circ$ circle.

Answer

$\begin{gathered}m\angle APD=64^\circ,\\m\angle YPD=116^\circ.\end{gathered}$

Key idea

The half-difference belongs to the specified rays toward the circle.
Problem 3 Three arcs and one exterior angle

Ray $PT$ is tangent at $T$ , and a secant has order $P,A,B$ . Near arc $TA$ excludes $B$ ; far arc $BT$ excludes $A$ . The near arc measures $52^\circ$ . Arc $AB$ excluding $T$ measures $124^\circ$ . Find the far arc and $m\angle TPA$ .

Schematic; use the stated givens.

Text description of this figure

PT is tangent at T. Secant ray order is P,A,B. Blue near arc TA excludes B; dashed red far arc BT excludes A. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  The three consecutive routes TA, AB, and BT fill the circle.
2. Hint 2
  After finding the far route, subtract the near route and halve the difference.
Answer

$\begin{gathered}m(\text{far arc }BT)=184^\circ,\\m\angle TPA=66^\circ.\end{gathered}$

Full solution

The known near arc and arc AB total $176^\circ$ .
Write f for the degree measure of the far arc.
Its complement gives
$f=360-176.$

The far arc is $184^\circ$ , a major route excluding A.
All three arc measures are positive and total $360^\circ$ .

The tangent-secant rule gives
$m\angle TPA=(184^\circ-52^\circ)/2.$

Thus the angle at P is $66^\circ$ .

Answer

$\begin{gathered}m(\text{far arc }BT)=184^\circ,\\m\angle TPA=66^\circ.\end{gathered}$

Key idea

An arc-completion step can come before the angle rule.
Problem 4 An obtuse angle between tangents

Rays $PT,PU$ are tangent to one circle at distinct $T,U$ , with $P$ strictly outside. Angle $TPU$ measures $108^\circ$ . Let $R$ lie on the major arc $TU$ . Find minor arc $TU$ and major arc $TRU$ . Explain why the obtuse angle is compatible with tangency.

Schematic; use the stated givens.

Text description of this figure

The rays PT and PU are tangent at T and U. Blue minor arc TU faces P; dashed red major arc TRU passes through R. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  The minor contact arc is supplementary to the angle between tangent rays.
2. Hint 2
  Subtract the minor arc from a full turn for the major arc. Check the base angles in triangle PTU.
Answer

$\begin{gathered}m(\widehat{TU})=72^\circ,\\m(\widehat{TRU})=288^\circ.\end{gathered}$

Full solution

The minor contact arc has measure
$m(\widehat{TU})=180^\circ-108^\circ.$

It is $72^\circ$ .
The remaining major arc is
$m(\widehat{TRU})=360^\circ-72^\circ.$

The major arc is $288^\circ$ .
Both arcs are positive and total $360^\circ$ .
Each base angle of triangle PTU is half the minor arc, or $36^\circ$ , leaving the given $108^\circ$ at P.
Tangency does not force an acute exterior angle.

Answer

$\begin{gathered}m(\widehat{TU})=72^\circ,\\m(\widehat{TRU})=288^\circ.\end{gathered}$

Key idea

A small minor contact arc gives an obtuse two-tangent angle.
Problem 5 Explain the inside half-sum

Points $A,B,C,D$ occur in that order around a circle. Chords $AC,BD$ cross at interior $P$ . Let $a$ be the degree measure of consecutive arc $AB$ and $c$ that of consecutive arc $CD$ . Prove $m\angle APB=(a+c)/2$ by joining $A$ to $D$ . Name the two inscribed angles and the triangle exterior angle you use.

Schematic; use the stated givens.

Text description of this figure

Chords AC and BD cross at interior P; blue arc AB and dashed red arc CD face vertical angles. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  P lies on both chords, so replace angle sides through P by rays to their circle endpoints.
2. Hint 2
  In triangle APD, PB extends PD. Identify the two remote interior angles.
Answer

$m\angle APB=(a+c)/2$ , using inscribed angles $CAD,BDA$ and the exterior angle $APB$ of triangle $APD$ .

Full solution

Since A,P,C are collinear in that order, angle PAD is angle CAD.
It intercepts consecutive arc CD, so
$m\angle PAD=c/2.$

Since D,P,B are collinear in that order, angle PDA is angle BDA.
It intercepts consecutive arc AB, so
$m\angle PDA=a/2.$

Ray PB is opposite PD, making APB an exterior angle of triangle APD.
The triangle exterior-angle theorem gives
$m\angle APB=a/2+c/2.$

Combining the fractions gives the required half-sum.
The argument uses distinct circle points with alternating chord endpoints and an intersection strictly inside.

The dotted auxiliary chord is used in the solution.

Answer

$m\angle APB=(a+c)/2$ , using inscribed angles $CAD,BDA$ and the exterior angle $APB$ of triangle $APD$ .

Key idea

An auxiliary chord turns a circle sum into a triangle exterior-angle sum.
Problem 6 Recover an algebraic arc and the rest

Points $A,B,C,D$ occur in that order around a circle. Chords $AC,BD$ cross at interior $P$ . The degree measures of consecutive arcs $AB,CD$ are $3x+14,5x+2$ . Angle $APB$ measures $(3x+22)^\circ$ . Consecutive arc $BC$ is $98^\circ$ .

Find $x$ , then find consecutive arc $DA$ . Verify all four arcs.

Schematic; use the stated givens.

Text description of this figure

Chords AC and BD cross at interior P; blue arc AB and dashed red arc CD face vertical angles. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  The inside angle is half the sum of arcs AB and CD.
2. Hint 2
  Solve that equation before using 360 degrees for all four consecutive arcs.
3. Hint 3
  Substitute into every arc expression to check positivity and the full-circle total.
Answer

$\begin{gathered}x=14,\\m(\text{arc }DA)=134^\circ.\end{gathered}$

Full solution

The two arc expressions add to $(8x+16)^\circ$ .
Doubling the angle expression gives $(6x+44)^\circ$ .
These quantities must be equal.

Simplifying gives
$6x+44=8x+16.$

Thus $x=14$ .
Arc AB is $56^\circ$ , and arc CD is $72^\circ$ .
Their half-sum is $64^\circ$ , matching $3x+22$ degrees.

The three known arcs total $226^\circ$ .
Writing d for the degree measure of arc DA gives
$d=360-226.$

The remaining arc is $134^\circ$ .
The consecutive arc measures are $56^\circ,98^\circ,72^\circ,134^\circ$ ; each is positive, and their sum is $360^\circ$ .

Answer

$\begin{gathered}x=14,\\m(\text{arc }DA)=134^\circ.\end{gathered}$

Key idea

Solve the angle relation, then test the whole arc configuration.
Problem 7 Keep both tangent rays distinct

Ray $PT$ is tangent at $T$ , and a secant has order $P,A,B$ . Near arc $TA$ excludes $B$ ; far arc $BT$ excludes $A$ . The near arc is $68^\circ$ , and angle $TPA$ is $47^\circ$ . Point $Q$ lies on the tangent line with order $P,T,Q$ .

Find arc $AB$ excluding $T$ .

Find $m\angle QTA$ and $m\angle PAT$ .

Schematic; use the stated givens.

Text description of this figure

PT is tangent at T. Secant ray order is P,A,B. Blue near arc TA excludes B; dashed red far arc BT excludes A. Dotted chord TA completes triangle PTA. Q is on the opposite tangent ray TQ. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  Use the exterior angle to recover the far arc BT.
2. Hint 2
  The tangent ray TP selects the near arc TA, but TQ is its opposite.
3. Hint 3
  At A, ray AP is opposite AB. Use the inscribed angle TAB before taking its supplement.
Answer

$\begin{gathered}m(\text{arc }AB)=130^\circ,\\m\angle QTA=146^\circ,\\m\angle PAT=99^\circ.\end{gathered}$

Full solution

The far arc has measure
$m(\text{far arc }BT)=2(47^\circ)+68^\circ.$

It is $162^\circ$ .
The remaining arc is
$m(\text{arc }AB)=360^\circ-68^\circ-162^\circ.$

Thus arc AB is $130^\circ$ , and the three positive arcs complete the circle.

The tangent-chord angle PTA selects the near arc and is $34^\circ$ .
Ray TQ is opposite TP, so
$m\angle QTA=180^\circ-34^\circ.$

Angle QTA is $146^\circ$ .
Inscribed angle TAB selects the far arc and is $81^\circ$ .
Since AP is opposite AB,
$m\angle PAT=180^\circ-81^\circ.$

Angle PAT is $99^\circ$ .
The triangle angles $47^\circ,34^\circ,99^\circ$ sum to $180^\circ$ .

Answer

$\begin{gathered}m(\text{arc }AB)=130^\circ,\\m\angle QTA=146^\circ,\\m\angle PAT=99^\circ.\end{gathered}$

Key idea

Name the tangent and secant ray branches before using supplementary angles.
Problem 8 Read the auxiliary triangle backward

Two secants from external $P$ have orders $P,A,B$ and $P,D,C$ . The circle order is $A,B,C,D$ . Far arc $BC$ excludes $A,D$ ; near arc $DA$ excludes $B,C$ . After joining $B$ to $D$ , the inscribed angles $ABD$ and $BDC$ measure $29^\circ$ and $97^\circ$ .

Find the near and far arc measures.

Find $m\angle APD$ and explain its triangle relationship.

Schematic; use the stated givens.

Text description of this figure

External P has secant ray orders P,A,B and P,D,C. Blue far arc BC excludes A,D; dashed red near arc DA excludes B,C. Dotted chord BD supplies a triangle. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  Double each inscribed angle to find its intercepted route.
2. Hint 2
  Angle BDC is exterior to triangle PBD; angle PBD is angle ABD.
Answer

$\begin{gathered}m(\text{near arc }DA)=58^\circ,\\m(\text{far arc }BC)=194^\circ,\\m\angle APD=68^\circ.\end{gathered}$

Full solution

Angle ABD intercepts near arc DA, giving
$m(\text{near arc }DA)=2(29^\circ).$

The near arc is $58^\circ$ .
Angle BDC intercepts far arc BC excluding D, giving
$m(\text{far arc }BC)=2(97^\circ).$

The far arc is $194^\circ$ .
It may be major.
The remaining two arcs have a positive total of $108^\circ$ ; each may be $54^\circ$ .

Because P,D,C are in that order, angle BDC is exterior to triangle PBD.
Hence
$97^\circ=m\angle APD+29^\circ.$

The angle APD is $68^\circ$ , agreeing with half the difference of 194 and 58 degrees.

Answer

$\begin{gathered}m(\text{near arc }DA)=58^\circ,\\m(\text{far arc }BC)=194^\circ,\\m\angle APD=68^\circ.\end{gathered}$

Key idea

The exterior circle rule can be recovered directly from its auxiliary triangle.
Problem 9 Transfer from a tangent angle to an inscribed angle

Rays $PT,PU$ are tangent to one circle at distinct $T,U$ , with $P$ strictly outside. Point $R$ lies on major arc $TU$ ; point $S$ lies on minor arc $TU$ . The minor arc measures $(3x+4)^\circ$ , and inscribed angle $TRU$ measures $(x+15)^\circ$ .

Find $x$ and $m\angle TPU$ .

Find $m\angle TSU$ , identifying its intercepted arc.

Schematic; use the stated givens.

Text description of this figure

S lies on minor arc TU. The rays PT and PU are tangent at T and U. Blue minor arc TU faces P; dashed red major arc TRU passes through R. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  R excludes the minor arc TU, so double angle TRU to get that arc.
2. Hint 2
  The tangent angle is supplementary to the minor contact arc.
3. Hint 3
  S is on the minor arc, so its intercepted arc is the major route through R.
Answer

$\begin{gathered}x=26,\\m\angle TPU=98^\circ,\\m\angle TSU=139^\circ.\end{gathered}$

Full solution

Angle TRU is half the minor arc TU.
Thus
$3x+4=2(x+15).$

Solving gives $x=26$ .
The minor arc is $82^\circ$ , and the major arc TRU is $278^\circ$ .
Both are positive and sum to $360^\circ$ .

The tangent angle is
$m\angle TPU=180^\circ-82^\circ.$

It measures $98^\circ$ .
Since S is on the minor arc, angle TSU intercepts the major arc TRU.
Therefore
$m\angle TSU=278^\circ/2.$

Angle TSU is $139^\circ$ .

Answer

$\begin{gathered}x=26,\\m\angle TPU=98^\circ,\\m\angle TSU=139^\circ.\end{gathered}$

Key idea

Changing the vertex from outside to on the circle changes the arc rule.
Problem 10 Reconstruct four linked arcs

Points $A,B,C,D$ occur in order around a circle. Chords $AC,BD$ meet at $X$ . Side lines $AB,DC$ meet at external $P$ with orders $P,A,B$ and $P,D,C$ . Far arc $BC$ excludes $A,D$ ; near arc $DA$ excludes $B,C$ . Angle $AXB$ measures $104^\circ$ , and angle $APD$ measures $28^\circ$ . Consecutive arc $AB$ measures $91^\circ$ .

Find all four consecutive arc measures $AB,BC,CD,DA$ . Verify positivity, the full-circle total, and both angle givens.

Schematic; use the stated givens.

Text description of this figure

External P has secant ray orders P,A,B and P,D,C. Blue far arc BC excludes A,D; dashed red near arc DA excludes B,C. Chords AC and BD meet at X. Schematic: incidence and point order are exact; use stated measures, not proportions.
1. Hint 1
  The inside angle fixes the sum of arcs AB and CD.
2. Hint 2
  Subtract that sum from 360 degrees to find the sum of BC and DA. Their difference comes from the exterior angle.
3. Hint 3
  Add the sum and difference equations to recover the far arc, then subtract to find the near arc.
Answer

$\begin{gathered}m(\text{arc }AB)=91^\circ,\\m(\text{arc }BC)=104^\circ,\\m(\text{arc }CD)=117^\circ,\\m(\text{arc }DA)=48^\circ.\end{gathered}$

Full solution

The inside angle gives
$m(\text{arc }AB)+m(\text{arc }CD)=208^\circ.$

Subtracting arc AB gives arc CD of $117^\circ$ .
Write b and d for the degree measures of far arc BC and near arc DA.
The full circle gives
$b+d=152.$

The exterior angle gives
$b-d=56.$

Adding these equations gives $2b=208$ , so $b=104$ .
Subtracting from 152 gives $d=48$ .

The four arc measures are positive, and
$91^\circ+104^\circ+117^\circ+48^\circ=360^\circ.$

The far arc exceeds the near arc, agreeing with the stated exterior ray orders.
The inside check is
$(91^\circ+117^\circ)/2=104^\circ.$

The exterior check is
$(104^\circ-48^\circ)/2=28^\circ.$

Answer

$\begin{gathered}m(\text{arc }AB)=91^\circ,\\m(\text{arc }BC)=104^\circ,\\m(\text{arc }CD)=117^\circ,\\m(\text{arc }DA)=48^\circ.\end{gathered}$

Key idea

A linked inside sum and outside difference determine the remaining arc pair.

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