Two radii form an isosceles triangle; find angles

4.MD.C.74.G.A.2 · take · grade 4

Archetype: Isosceles and Equilateral Angle Chaining · step in a 6-type progression

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

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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.

\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.

\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.

\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^\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. The value a = 45 is a sensible mid-size angle for a vertex on the circle, 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!