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Inscribed Angles: 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 Exclude the vertex

Points $A,V,B,W$ occur in that order on a circle. Arc $AVB$ measures $112^\circ$ .

Name the arc intercepted by $\angle AVB$ , find its measure, and find $m\angle AVB$ .

The unknown angle is at V.

Text description of this figure

V lies on arc AVB and W on the other arc AB. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  The intercepted arc cannot contain the angle's vertex.
2. Hint 2
  The two arcs from $A$ to $B$ total $360^\circ$ . Halve the one away from $V$ .
Answer

Intercepted arc $AWB$ .
$\begin{gathered}m(\widehat{AWB})=248^\circ,\\m\angle AVB=124^\circ.\end{gathered}$

Full solution

The arc from $A$ through $W$ to $B$ excludes $V$ , so it is the intercepted arc.

Subtract the arc containing $V$ from a full circle.
$m(\widehat{AWB})=360^\circ-112^\circ.$

The intercepted arc measures $248^\circ$ .
The angle is half that arc.
$m\angle AVB=248^\circ/2.$

The angle measures $124^\circ$ .

Answer

Intercepted arc $AWB$ .
$\begin{gathered}m(\widehat{AWB})=248^\circ,\\m\angle AVB=124^\circ.\end{gathered}$

Key idea

Choose the arc by the vertex position before doing arithmetic.
Problem 2 Center on a side

$VA$ is a diameter with center $O$ . Point $B$ is another point on the circle, and $m\angle OVB=27^\circ$ .

Find $m\angle VOB$ and the measure of arc $BA$ excluding $V$ . Justify the steps using the equal radii and the diameter.

The diameter is given; the central angles are unknown.

Text description of this figure

The center O lies on side VA of angle AVB; VA is a diameter. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Compare the two base angles of triangle $VOB$ .
2. Hint 2
  Find the angle at $O$ in that triangle, then use opposite rays $OV$ and $OA$ .
Answer

$\begin{gathered}m\angle VOB=126^\circ,\\m(\widehat{BA})=54^\circ.\end{gathered}$

Full solution

Radii $OV$ and $OB$ are equal, so the angle $VBO$ also measures $27^\circ$ .

The triangle sum gives the first center angle.
$m\angle VOB=180^\circ-27^\circ-27^\circ.$

The center angle $VOB$ measures $126^\circ$ .
Rays $OV$ and $OA$ are opposite.
$m\angle BOA=180^\circ-126^\circ.$

Angle $BOA$ measures $54^\circ$ and is the central angle of arc $BA$ excluding $V$ , so that arc measures $54^\circ$ .

Answer

$\begin{gathered}m\angle VOB=126^\circ,\\m(\widehat{BA})=54^\circ.\end{gathered}$

Key idea

Equal radii and a straight angle establish the diameter-side case.
Problem 3 Rebuild the outside case

In a circle with center $O$ , $VD$ is a diameter outside $\angle AVB$ . Ray $VA$ lies between rays $VD$ and $VB$ . The arc order away from $V$ is $D,A,B$ .

Arc $DA$ measures $u$ and arc $DB$ measures $v$ , with $0<u<v<180^\circ$ . Using the diameter-side case, prove that $m\angle AVB=(v-u)/2$ . State which arc this angle intercepts.

Diameter VD and rays VA and VB are shown in the stated order.

Text description of this figure

Diameter VD lies outside angle AVB. The arc order away from V is D, A, B. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Apply the side case to $\angle DVA$ and $\angle DVB$ separately.
2. Hint 2
  Subtract the smaller angle from the larger, and identify the corresponding arc subtraction.
Answer

$m\angle AVB=(v-u)/2$ , intercepting arc $AB$ away from $V$ .

Full solution

The diameter-side case applies to the two angles with a side along $VD$ .
$m\angle DVA=u/2.$

For the larger angle it gives
$m\angle DVB=v/2.$

Ray $VA$ is between $VD$ and $VB$ , so subtract the smaller angle.
$m\angle AVB=v/2-u/2.$

Combine the fractions.
$m\angle AVB=(v-u)/2.$

The order $D,A,B$ makes arc $AB$ the remainder after arc $DA$ is removed from arc $DB$ .
Its measure is $v-u$ , and it excludes $V$ , so it is the intercepted arc.

Answer

$m\angle AVB=(v-u)/2$ , intercepting arc $AB$ away from $V$ .

Key idea

An outside auxiliary diameter requires matching angle and arc subtraction.
Problem 4 Center inside the angle

Diameter $VD$ lies inside $\angle AVB$ . Along the intercepted arc the order is $A,D,B$ . Arc $AD$ measures $66^\circ$ , and arc $DB$ measures $92^\circ$ .

Find $m\angle AVD$ , $m\angle DVB$ , and $m\angle AVB$ . Explain why the angle parts add.

The highlighted arc pieces are the supplied givens.

Text description of this figure

Diameter VD splits angle AVB and intercepted arc ADB. Solid arc AD and dashed arc DB add. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Use the side case for each angle with a side along $VD$ .
2. Hint 2
  Ray $VD$ lies between $VA$ and $VB$ , so the whole angle is the sum of the two parts.
Answer

$\begin{gathered}m\angle AVD=33^\circ,\\m\angle DVB=46^\circ,\\m\angle AVB=79^\circ.\end{gathered}$

Full solution

The diameter-side case gives the first part.
$m\angle AVD=66^\circ/2.$

The first part measures $33^\circ$ .
The same case gives the second part.
$m\angle DVB=92^\circ/2.$

The second part measures $46^\circ$ .
Ray $VD$ is inside the angle, so the adjacent parts add.
$m\angle AVB=33^\circ+46^\circ.$

The whole angle measures $79^\circ$ , half the full intercepted arc of $158^\circ$ .

Answer

$\begin{gathered}m\angle AVD=33^\circ,\\m\angle DVB=46^\circ,\\m\angle AVB=79^\circ.\end{gathered}$

Key idea

An inside diameter splits the angle and its intercepted arc into matching parts.
Problem 5 Equal angles, then the other arc

Points $P,Q$ lie on the same open arc between $A,B$ . Point $R$ lies on the other open arc.

Angle $APB$ measures $(3x+5)^\circ$ , and $AQB$ measures $(5x-21)^\circ$ . Find $x$ , $m\angle APB$ , and $m\angle ARB$ .

Each vertex excludes the arc on the other side of chord AB.

Text description of this figure

P and Q lie on major arc AB; R lies on minor arc AB. The three vertices exclude different arcs. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Compare the arcs intercepted by the angles at $P$ and $Q$ .
2. Hint 2
  Set their angle expressions equal. Then select the other arc for the vertex $R$ .
Answer

$\begin{gathered}x=13,\\m\angle APB=44^\circ,\\m\angle ARB=136^\circ.\end{gathered}$

Full solution

The angles at $P,Q$ intercept the same arc and are equal.
$3x+5=5x-21.$

Subtract $3x$ and add $21$ .
$26=2x.$

Divide by $2$ .
$x=13.$

Substitute into the angle at $P$ .
$m\angle APB=(3(13)+5)^\circ.$

The angle at $P$ measures $44^\circ$ .
The vertex $R$ selects the other arc.
$m\angle ARB=180^\circ-44^\circ.$

The angle at $R$ measures $136^\circ$ .

Answer

$\begin{gathered}x=13,\\m\angle APB=44^\circ,\\m\angle ARB=136^\circ.\end{gathered}$

Key idea

Same-arc equality can determine a variable before an opposite-arc angle is found.
Problem 6 A triangle on a diameter

Triangle $ACB$ is inscribed in a circle with diameter $AB$ . Its angle at $A$ measures $(2x+7)^\circ$ , and its angle at $B$ measures $(3x-2)^\circ$ .

Find $x$ and all three triangle angles.

AB is given as a diameter; use the angle expressions, not the sketch.

Text description of this figure

AB is a diameter with center O, and C is another point on the circle. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  A diameter fixes the angle at the third point on the circle.
2. Hint 2
  The other two angles must total $90^\circ$ .
Answer

$\begin{gathered}x=17,\\m\angle CAB=41^\circ,\\m\angle ABC=49^\circ,\\m\angle ACB=90^\circ.\end{gathered}$

Full solution

The angle at $C$ intercepts a semicircle and measures $90^\circ$ .

The other two triangle angles total $90^\circ$ .
$(2x+7)+(3x-2)=90.$

Collect terms.
$5x+5=90.$

Subtract $5$ and divide by $5$ .
$x=17.$

Substitute to find the angle at $A$ .
$m\angle CAB=(2(17)+7)^\circ.$

The angle at $A$ is $41^\circ$ .
Substitution at $B$ gives
$m\angle ABC=(3(17)-2)^\circ.$

The angle at $B$ is $49^\circ$ .
These angles total $90^\circ$ , completing the triangle sum with the angle at $C$ .

Answer

$\begin{gathered}x=17,\\m\angle CAB=41^\circ,\\m\angle ABC=49^\circ,\\m\angle ACB=90^\circ.\end{gathered}$

Key idea

Use the semicircle theorem before solving the angle equation.
Problem 7 Recognize a diameter

Distinct points $A,B,C$ lie on a circle with center $O$ . Angle $CAB$ measures $42^\circ$ , and $CBA$ measures $48^\circ$ .

Determine whether segment $AB$ passes through $O$ . Give a complete geometric justification.
1. Hint 1
  Find the angle at $C$ from the triangle sum.
2. Hint 2
  What does the intercepted arc measure when an inscribed angle is right?
Answer

Yes. Segment $AB$ is a diameter and passes through $O$ .

Full solution

The triangle sum determines the remaining angle.
$m\angle ACB=180^\circ-42^\circ-48^\circ.$

This angle is $90^\circ$ , so its intercepted arc $AB$ excluding $C$ measures $180^\circ$ .

By the converse of the semicircle theorem, $AB$ is a diameter of this circle.
Every diameter passes through its center $O$ .

The solution identifies AB as a diameter after proving the right angle at C.

Answer

Yes. Segment $AB$ is a diameter and passes through $O$ .

Key idea

A proved right angle can establish a diameter by the converse.
Problem 8 Two tangent rays, two arcs

A circle has center $O$ . Points $T,A$ lie on it, and $m\angle OAT=34^\circ$ . Rays $TR,TS$ are opposite tangent rays at $T$ . Ray $TR$ is on the same side of line $OT$ as $A$ .

Find $m\angle RTA$ and $m\angle STA$ .

Find the minor and major arc measures from $T$ to $A$ .

The right-angle mark is for the tangent and radius; the tangent-chord angles are unknown.

Text description of this figure

TR and TS are opposite tangent rays at T. R and A lie on the same side of line OT. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Equal radii fix the angle $OTA$ . Compare it with the right angle between $TO$ and $TR$ .
2. Hint 2
  The opposite tangent angles are supplementary. Double each one for its selected arc.
Answer

$\begin{gathered}m\angle RTA=56^\circ,\\m\angle STA=124^\circ,\\m(\text{minor arc }TA)=112^\circ,\\m(\text{major arc }TA)=248^\circ.\end{gathered}$

Full solution

Equal radii $OT,OA$ make triangle $OTA$ isosceles, so the angle $OTA$ is $34^\circ$ .

The tangent is perpendicular to the radius at $T$ .
Subtract inside the right angle.
$m\angle RTA=90^\circ-34^\circ.$

The first tangent angle is $56^\circ$ .
The opposite ray gives the supplement.
$m\angle STA=180^\circ-56^\circ.$

The second angle is $124^\circ$ .
Double the first angle for the minor arc.
$m(\text{minor arc }TA)=2(56^\circ).$

The minor arc measures $112^\circ$ .
Double the second angle for the major arc.
$m(\text{major arc }TA)=2(124^\circ).$

The major arc measures $248^\circ$ .
The two arcs total $360^\circ$ , as required.

Answer

$\begin{gathered}m\angle RTA=56^\circ,\\m\angle STA=124^\circ,\\m(\text{minor arc }TA)=112^\circ,\\m(\text{major arc }TA)=248^\circ.\end{gathered}$

Key idea

An obtuse tangent-chord angle belongs to the major arc.
Problem 9 Transfer to tangent rays

A circle has center $O$ . Point $B$ lies on major arc $TA$ , and point $C$ lies on minor arc $TA$ . The inscribed angle $TBA$ measures $63^\circ$ .

At $T$ , ray $TR$ is tangent on the same side of line $OT$ as $A$ , and $TS$ is its opposite ray. Find $m\angle RTA$ , $m\angle STA$ , and $m\angle TCA$ .

Follow each vertex or tangent ray to its selected arc.

Text description of this figure

B lies on major arc TA and C on minor arc TA. TR and TS are opposite tangent rays. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  Recover the minor arc from the angle at $B$ .
2. Hint 2
  Ray $TR$ and vertex $B$ select the same arc. Ray $TS$ and vertex $C$ select the other.
Answer

$\begin{gathered}m\angle RTA=63^\circ,\\m\angle STA=117^\circ,\\m\angle TCA=117^\circ.\end{gathered}$

Full solution

Vertex $B$ excludes the minor arc $TA$ , so double its angle.
$m(\widehat{TA})=2(63^\circ).$

The minor arc measures $126^\circ$ , and the remaining major arc measures $234^\circ$ .

Ray $TR$ selects the minor arc.
$m\angle RTA=126^\circ/2.$

That angle measures $63^\circ$ .
Ray $TS$ selects the major arc.
$m\angle STA=234^\circ/2.$

That angle measures $117^\circ$ .
Vertex $C$ is on the minor arc and excludes it, also selecting the major arc.
$m\angle TCA=234^\circ/2.$

The angle at $C$ therefore measures $117^\circ$ .

Answer

$\begin{gathered}m\angle RTA=63^\circ,\\m\angle STA=117^\circ,\\m\angle TCA=117^\circ.\end{gathered}$

Key idea

Different kinds of angles agree when they select the same arc.
Problem 10 Linked arc and ray choices

A circle has center $O$ and distinct points $A,B,P,Q$ . Point $P$ lies on major arc $AB$ , and $Q$ on minor arc $AB$ .

Angle $APB$ measures $(4x+6)^\circ$ , and $AQB$ measures $(7x-13)^\circ$ .

At $A$ , ray $AS$ is tangent on the opposite side of line $OA$ from $B$ . Find $x$ , the measure of arc $AQB$ , and $m\angle SAB$ .

The tangent angle is unknown; no computed angle is shown.

Text description of this figure

At A the tangent ray AS is opposite B across line OA. P lies on major arc AB and Q on minor arc AB. Exact angle and arc measures come from the statement; the schematic preserves the stated incidence and point order.
1. Hint 1
  The two inscribed angles exclude the two arcs that together make a full circle.
2. Hint 2
  Use the angle sum to find $x$ . Vertex $P$ selects the minor arc; ray $AS$ selects the major arc.
Answer

$\begin{gathered}x=17,\\m(\widehat{AQB})=148^\circ,\\m\angle SAB=106^\circ.\end{gathered}$

Full solution

The intercepted arcs total $360^\circ$ , so their half-sized angles total $180^\circ$ .
$(4x+6)+(7x-13)=180.$

Collect terms.
$11x-7=180.$

Add $7$ and divide by $11$ .
$x=17.$

The angle at $P$ measures $74^\circ$ .
It intercepts minor arc $AQB$ .
$m(\widehat{AQB})=2(74^\circ).$

The minor arc measures $148^\circ$ and the remaining major arc measures $212^\circ$ .

Ray $AS$ is opposite $B$ 's side of the radius line and selects that major arc.
$m\angle SAB=212^\circ/2.$

The tangent angle measures $106^\circ$ .
The inscribed angle at $Q$ has that same measure, confirming the shared major arc.

Answer

$\begin{gathered}x=17,\\m(\widehat{AQB})=148^\circ,\\m\angle SAB=106^\circ.\end{gathered}$

Key idea

Transfer through the selected arc, keeping the vertex and tangent ray distinct.

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