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Two radii form an isosceles triangle; find angles · 10 practice problems

4.G.A.24.MD.C.7

Generated variants — 10

Freshly produced from the archetype’s parameters — problem, figure, and solution derived together.

Variant 1 answer: 50 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 18° 22° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 18 degrees and angle OBC = 22 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 18 degrees
  • Angle OBC = 22 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 18 degrees.
∠OAB=∠OBA=18∘\angle OAB = \angle OBA = 18^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 22 degrees.
∠OCB=∠OBC=22∘\angle OCB = \angle OBC = 22^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 18 + 22 = 40 degrees; angle at A = angle BAO + angle OAC = 18 + a; angle at C = angle BCO + angle OCA = 22 + a. Their sum is 180 degrees: 40 + (18 + a) + (22 + a) = 180, so 80 + 2a = 180, giving 2a = 100 and a = 50 degrees.
40∘+(18∘+a)+(22∘+a)=180∘⇒2a=100∘⇒a=50∘40^\circ + (18^\circ + a) + (22^\circ + a) = 180^\circ \Rightarrow 2a = 100^\circ \Rightarrow a = 50^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 50 degrees

Review

With a = 50 degrees, triangle ABC has angles 40 (at B), 68 (at A = 18+50), and 72 (at C = 22+50), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 2 answer: 60 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 12° 18° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 12 degrees and angle OBC = 18 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 12 degrees
  • Angle OBC = 18 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 12 degrees.
∠OAB=∠OBA=12∘\angle OAB = \angle OBA = 12^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 18 degrees.
∠OCB=∠OBC=18∘\angle OCB = \angle OBC = 18^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 12 + 18 = 30 degrees; angle at A = angle BAO + angle OAC = 12 + a; angle at C = angle BCO + angle OCA = 18 + a. Their sum is 180 degrees: 30 + (12 + a) + (18 + a) = 180, so 60 + 2a = 180, giving 2a = 120 and a = 60 degrees.
30∘+(12∘+a)+(18∘+a)=180∘⇒2a=120∘⇒a=60∘30^\circ + (12^\circ + a) + (18^\circ + a) = 180^\circ \Rightarrow 2a = 120^\circ \Rightarrow a = 60^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 60 degrees

Review

With a = 60 degrees, triangle ABC has angles 30 (at B), 72 (at A = 12+60), and 78 (at C = 18+60), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 3 answer: 45 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 30° 15° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 30 degrees and angle OBC = 15 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 30 degrees
  • Angle OBC = 15 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 30 degrees.
∠OAB=∠OBA=30∘\angle OAB = \angle OBA = 30^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 15 degrees.
∠OCB=∠OBC=15∘\angle OCB = \angle OBC = 15^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 30 + 15 = 45 degrees; angle at A = angle BAO + angle OAC = 30 + a; angle at C = angle BCO + angle OCA = 15 + a. Their sum is 180 degrees: 45 + (30 + a) + (15 + a) = 180, so 90 + 2a = 180, giving 2a = 90 and a = 45 degrees.
45∘+(30∘+a)+(15∘+a)=180∘⇒2a=90∘⇒a=45∘45^\circ + (30^\circ + a) + (15^\circ + a) = 180^\circ \Rightarrow 2a = 90^\circ \Rightarrow a = 45^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 45 degrees

Review

With a = 45 degrees, triangle ABC has angles 45 (at B), 75 (at A = 30+45), and 60 (at C = 15+45), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 4 answer: 20 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 40° 30° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 40 degrees and angle OBC = 30 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 40 degrees
  • Angle OBC = 30 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 40 degrees.
∠OAB=∠OBA=40∘\angle OAB = \angle OBA = 40^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 30 degrees.
∠OCB=∠OBC=30∘\angle OCB = \angle OBC = 30^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 40 + 30 = 70 degrees; angle at A = angle BAO + angle OAC = 40 + a; angle at C = angle BCO + angle OCA = 30 + a. Their sum is 180 degrees: 70 + (40 + a) + (30 + a) = 180, so 140 + 2a = 180, giving 2a = 40 and a = 20 degrees.
70∘+(40∘+a)+(30∘+a)=180∘⇒2a=40∘⇒a=20∘70^\circ + (40^\circ + a) + (30^\circ + a) = 180^\circ \Rightarrow 2a = 40^\circ \Rightarrow a = 20^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 20 degrees

Review

With a = 20 degrees, triangle ABC has angles 70 (at B), 60 (at A = 40+20), and 50 (at C = 30+20), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 5 answer: 45 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 15° 30° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 15 degrees and angle OBC = 30 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 15 degrees
  • Angle OBC = 30 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 15 degrees.
∠OAB=∠OBA=15∘\angle OAB = \angle OBA = 15^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 30 degrees.
∠OCB=∠OBC=30∘\angle OCB = \angle OBC = 30^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 15 + 30 = 45 degrees; angle at A = angle BAO + angle OAC = 15 + a; angle at C = angle BCO + angle OCA = 30 + a. Their sum is 180 degrees: 45 + (15 + a) + (30 + a) = 180, so 90 + 2a = 180, giving 2a = 90 and a = 45 degrees.
45∘+(15∘+a)+(30∘+a)=180∘⇒2a=90∘⇒a=45∘45^\circ + (15^\circ + a) + (30^\circ + a) = 180^\circ \Rightarrow 2a = 90^\circ \Rightarrow a = 45^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 45 degrees

Review

With a = 45 degrees, triangle ABC has angles 45 (at B), 60 (at A = 15+45), and 75 (at C = 30+45), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 6 answer: 40 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 10° 40° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 10 degrees and angle OBC = 40 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 10 degrees
  • Angle OBC = 40 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 10 degrees.
∠OAB=∠OBA=10∘\angle OAB = \angle OBA = 10^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 40 degrees.
∠OCB=∠OBC=40∘\angle OCB = \angle OBC = 40^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 10 + 40 = 50 degrees; angle at A = angle BAO + angle OAC = 10 + a; angle at C = angle BCO + angle OCA = 40 + a. Their sum is 180 degrees: 50 + (10 + a) + (40 + a) = 180, so 100 + 2a = 180, giving 2a = 80 and a = 40 degrees.
50∘+(10∘+a)+(40∘+a)=180∘⇒2a=80∘⇒a=40∘50^\circ + (10^\circ + a) + (40^\circ + a) = 180^\circ \Rightarrow 2a = 80^\circ \Rightarrow a = 40^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 40 degrees

Review

With a = 40 degrees, triangle ABC has angles 50 (at B), 50 (at A = 10+40), and 80 (at C = 40+40), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 7 answer: 30 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 35° 25° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 35 degrees and angle OBC = 25 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 35 degrees
  • Angle OBC = 25 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 35 degrees.
∠OAB=∠OBA=35∘\angle OAB = \angle OBA = 35^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 25 degrees.
∠OCB=∠OBC=25∘\angle OCB = \angle OBC = 25^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 35 + 25 = 60 degrees; angle at A = angle BAO + angle OAC = 35 + a; angle at C = angle BCO + angle OCA = 25 + a. Their sum is 180 degrees: 60 + (35 + a) + (25 + a) = 180, so 120 + 2a = 180, giving 2a = 60 and a = 30 degrees.
60∘+(35∘+a)+(25∘+a)=180∘⇒2a=60∘⇒a=30∘60^\circ + (35^\circ + a) + (25^\circ + a) = 180^\circ \Rightarrow 2a = 60^\circ \Rightarrow a = 30^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 30 degrees

Review

With a = 30 degrees, triangle ABC has angles 60 (at B), 65 (at A = 35+30), and 55 (at C = 25+30), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 8 answer: 48 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 28° 14° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 28 degrees and angle OBC = 14 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 28 degrees
  • Angle OBC = 14 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 28 degrees.
∠OAB=∠OBA=28∘\angle OAB = \angle OBA = 28^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 14 degrees.
∠OCB=∠OBC=14∘\angle OCB = \angle OBC = 14^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 28 + 14 = 42 degrees; angle at A = angle BAO + angle OAC = 28 + a; angle at C = angle BCO + angle OCA = 14 + a. Their sum is 180 degrees: 42 + (28 + a) + (14 + a) = 180, so 84 + 2a = 180, giving 2a = 96 and a = 48 degrees.
42∘+(28∘+a)+(14∘+a)=180∘⇒2a=96∘⇒a=48∘42^\circ + (28^\circ + a) + (14^\circ + a) = 180^\circ \Rightarrow 2a = 96^\circ \Rightarrow a = 48^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 48 degrees

Review

With a = 48 degrees, triangle ABC has angles 42 (at B), 76 (at A = 28+48), and 62 (at C = 14+48), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 9 answer: 45 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 20° 25° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 20 degrees and angle OBC = 25 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 20 degrees
  • Angle OBC = 25 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 20 degrees.
∠OAB=∠OBA=20∘\angle OAB = \angle OBA = 20^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 25 degrees.
∠OCB=∠OBC=25∘\angle OCB = \angle OBC = 25^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 20 + 25 = 45 degrees; angle at A = angle BAO + angle OAC = 20 + a; angle at C = angle BCO + angle OCA = 25 + a. Their sum is 180 degrees: 45 + (20 + a) + (25 + a) = 180, so 90 + 2a = 180, giving 2a = 90 and a = 45 degrees.
45∘+(20∘+a)+(25∘+a)=180∘⇒2a=90∘⇒a=45∘45^\circ + (20^\circ + a) + (25^\circ + a) = 180^\circ \Rightarrow 2a = 90^\circ \Rightarrow a = 45^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 45 degrees

Review

With a = 45 degrees, triangle ABC has angles 45 (at B), 65 (at A = 20+45), and 70 (at C = 25+45), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!
Variant 10 answer: 45 degrees

In the figure below, find the measure of ∠a\angle a. (Point OO is the center of the circle.)

A B C O 25° 20° a
Show solution

Understand

Points A, B, C lie on a circle with center O. The radii OA, OB, OC and the chords are drawn. At B, angle OBA = 25 degrees and angle OBC = 20 degrees. I must find angle a, which is angle OAC at vertex A.

Givens
  • O is the center, so OA, OB, OC are all radii of equal length
  • Angle OBA = 25 degrees
  • Angle OBC = 20 degrees
  • Angle a = angle OAC
Unknowns
  • The measure of angle a (= angle OAC)
Constraints
  • All radii of one circle are equal, so each triangle with two radii is isosceles
  • An isosceles triangle has two equal base angles
  • Angles in a triangle sum to 180 degrees

Plan

#7 Identify Subproblems · also uses: #1 Draw a Diagram#13 Convert to Algebra

The radii cut the figure into three isosceles triangles, so I treat each as a subproblem. Using equal base angles I express the angles of the big triangle ABC in terms of a, then set their sum to 180 degrees to solve for a.

Execute

#7 Identify Subproblems 4.G.A.2
OA and OB are both radii, so OA = OB and triangle OAB is isosceles. Its base angles are equal, so angle OAB = angle OBA = 25 degrees.
∠OAB=∠OBA=25∘\angle OAB = \angle OBA = 25^\circ
Two radii are equal, so the triangle they form is isosceles with equal base angles.
#7 Identify Subproblems 4.G.A.2
OB and OC are radii, so OB = OC and triangle OBC is isosceles. Its base angles are equal, so angle OCB = angle OBC = 20 degrees.
∠OCB=∠OBC=20∘\angle OCB = \angle OBC = 20^\circ
The same equal-radius idea makes the second triangle isosceles too.
#7 Identify Subproblems 4.G.A.2
OA and OC are radii, so OA = OC and triangle OAC is isosceles. Its base angles are equal, so angle OCA = angle OAC = a.
∠OCA=∠OAC=a\angle OCA = \angle OAC = a
The third pair of radii forms one more isosceles triangle with two equal angles, each called a.
#13 Convert to Algebra 4.MD.C.7
In big triangle ABC: angle at B = 25 + 20 = 45 degrees; angle at A = angle BAO + angle OAC = 25 + a; angle at C = angle BCO + angle OCA = 20 + a. Their sum is 180 degrees: 45 + (25 + a) + (20 + a) = 180, so 90 + 2a = 180, giving 2a = 90 and a = 45 degrees.
45∘+(25∘+a)+(20∘+a)=180∘⇒2a=90∘⇒a=45∘45^\circ + (25^\circ + a) + (20^\circ + a) = 180^\circ \Rightarrow 2a = 90^\circ \Rightarrow a = 45^\circ
Writing each big-triangle angle as a known part plus a, then using the 180-degree total, pins down a.
Answer: 45 degrees

Review

With a = 45 degrees, triangle ABC has angles 45 (at B), 70 (at A = 25+45), and 65 (at C = 20+45), which total 180 degrees, so it checks out.

Use the central-angle viewpoint (tool 15): find each isosceles triangle's apex angle at O, add them to get the full central angle, and relate it back to angle a as an organized cross-check.

Standards · min grade 4

  • 4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines — Recognizing that two radii form an isosceles triangle with equal base angles.
  • 4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Adding the part-angles and using the triangle-sum to solve for angle a.
💡 This only needs Grade 4 angle-adding plus knowing two equal radii make an isosceles triangle!