Young Math Brains

Inscribed angle trap: choosing the correct reflex central angle

7 min read

Question

A, B and D lie on a circle with centre O. If ∠ADB = 130°, find ∠OAB.

The Circles question most students fail (is it 60° or 120°?)Watch on YouTube

We need to find ∠OAB\angle OAB by tracing the arc intercepted by the given angle, distinguishing the reflex and minor central angles, and then using the equal radii in triangle AOB. Each value is derived in that order below.

ABDO∠ADB = 130°intercepted major arc ABminor ∠AOB
The 130-degree angle at D faces the long, highlighted arc. Doubling 130 therefore gives the reflex central angle, not the smaller angle inside triangle AOB.

Step 1: identify the arc intercepted by angle ADB

The arms of the angle, DADA and DBDB, intersect the circle at points AA and BB. The intercepted arc is the arc ABAB that does not contain the vertex DD. Since DD lies on the minor arc ABAB, the intercepted arc is the major arc ABAB.

∠ADB=130∘(Given inscribed angle at vertex D)\angle ADB = 130^\circ \quad (\text{Given inscribed angle at vertex } D)
Vertex D lies on minor arc AB  ⟹  Intercepted arc=major arc AB\text{Vertex } D \text{ lies on minor arc } AB \implies \text{Intercepted arc} = \text{major arc } AB

Step 2: calculate the reflex central angle

Apply the Inscribed Angle Theorem: the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Reflex ∠AOB=2×∠ADB[Central angle is double the inscribed angle for major arc AB]\text{Reflex } \angle AOB = 2 \times \angle ADB \quad [\text{Central angle is double the inscribed angle for major arc } AB]
Reflex ∠AOB=2×130∘\text{Reflex } \angle AOB = 2 \times 130^\circ
Reflex ∠AOB=260∘\text{Reflex } \angle AOB = 260^\circ

Step 3: find the interior central angle inside triangle AOB

The angle inside △AOB\triangle AOB is the minor central angle. The sum of the reflex angle and minor angle around the centre OO is a complete angle of 360∘360^\circ:

Minor ∠AOB+Reflex ∠AOB=360∘(Complete angle around centre O)\text{Minor } \angle AOB + \text{Reflex } \angle AOB = 360^\circ \quad (\text{Complete angle around centre } O)
Minor ∠AOB+260∘=360∘\text{Minor } \angle AOB + 260^\circ = 360^\circ
Subtract 260∘ from both sides: Minor ∠AOB=360∘−260∘\text{Subtract } 260^\circ \text{ from both sides: } \text{Minor } \angle AOB = 360^\circ - 260^\circ
Minor ∠AOB=100∘\text{Minor } \angle AOB = 100^\circ

Step 4: use the isosceles triangle AOB

Since OAOA and OBOB are radii of the circle, OA=OBOA = OB. In an isosceles triangle, angles opposite to equal sides are equal. Let ∠OAB=∠OBA=θ\angle OAB = \angle OBA = \theta:

In △AOB:OA=OB(Radii of the same circle)\text{In } \triangle AOB: OA = OB \quad (\text{Radii of the same circle})
  ⟹  ∠OAB=∠OBA=θ(Angles opposite to equal sides are equal)\implies \angle OAB = \angle OBA = \theta \quad (\text{Angles opposite to equal sides are equal})
∠OAB+∠OBA+∠AOB=180∘(Angle sum property of △AOB)\angle OAB + \angle OBA + \angle AOB = 180^\circ \quad (\text{Angle sum property of } \triangle AOB)
θ+θ+100∘=180∘\theta + \theta + 100^\circ = 180^\circ
2θ+100∘=180∘2\theta + 100^\circ = 180^\circ
2θ=180∘−100∘[Subtract 100∘ from both sides]2\theta = 180^\circ - 100^\circ \quad [\text{Subtract } 100^\circ \text{ from both sides]}
2θ=80∘2\theta = 80^\circ
θ=80∘2[Divide both sides by 2]\theta = \frac{80^\circ}{2} \quad [\text{Divide both sides by } 2]
θ=40∘\theta = 40^\circ
  ⟹  ∠OAB=40∘\implies \angle OAB = 40^\circ

Step 5: verify the three angles of triangle AOB

∠OAB+∠OBA+∠AOB=180∘\angle OAB + \angle OBA + \angle AOB = 180^\circ
40∘+40∘+100∘=180∘40^\circ + 40^\circ + 100^\circ = 180^\circ
80∘+100∘=180∘80^\circ + 100^\circ = 180^\circ
180∘=180∘(Angle sum verified ✓)180^\circ = 180^\circ \quad (\text{Angle sum verified } \checkmark)

Four checks for major-arc questions

  • Trace the arms of the inscribed angle. They identify the two arc endpoints.
  • Exclude the vertex from the intercepted arc. The required arc is on the opposite side.
  • Label reflex and minor explicitly. Do not write only “angle AOB” when two values are possible.
  • Use the size test. An obtuse angle at the circumference must correspond to a major arc greater than 180 degrees.

Final step: state the verified angle

∠OAB=40∘\boxed{\angle OAB = 40^\circ}

Young Math Brains makes you select the intercepted arc before applying the central-angle theorem, so the 260-versus-100 distinction becomes visible.

Practise major and minor arcs

Your turn

Which angle belongs inside triangle AOB: the 260-degree reflex angle or the 100-degree minor angle?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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