Tangents Inclined at 60°: Prove OP = 2r | Class 10 Maths

Quick answer: If two tangents PA and PB from an external point P to a circle (centre O, radius r) are inclined at 60°, then OP bisects the angle, so ∠APO = 30°. In right triangle OAP, $\sin 30^\circ = \frac{r}{OP}$ gives OP = 2r, and $\tan 30^\circ = \frac{r}{PA}$ gives PA = PB = r√3. Triangle PAB is equilateral, so the chord AB is also r√3, and ∠AOB = 120°.

The setup

From an external point P, two tangents PA and PB are drawn to a circle with centre O and radius r, and ∠APB = 60°.

        A
       /|
      / | r
     /30|
  P ----+---- O
     \30|
      \ | r
       \|
        B
 ∠OAP = ∠OBP = 90°

Two facts from the Circles chapter are used:

• The radius to the point of contact is perpendicular to the tangent, so ∠OAP = ∠OBP = 90°.
• Tangents from an external point are equal, so PA = PB.

Why OP bisects ∠APB: Triangles OAP and OBP are congruent by RHS (OA = OB = r, OP is common, right angles at A and B). So ∠APO = ∠BPO = 60° ÷ 2 = 30°.

Result 1: OP = 2r

Step 1: In right triangle OAP, OA is opposite ∠APO and OP is the hypotenuse.

$\sin 30^\circ = \frac{OA}{OP} = \frac{r}{OP}$

Step 2: Use $\sin 30^\circ = \frac{1}{2}$.

$\frac{1}{2} = \frac{r}{OP}$

$OP = 2r$

Hence proved.

Result 2: tangent length PA = r√3

Step 1: In right triangle OAP:

$\tan 30^\circ = \frac{OA}{PA} = \frac{r}{PA}$

Step 2: Use $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

$PA = r\sqrt{3}$

Since PA = PB, both tangents have length $r\sqrt{3}$.

Check with Pythagoras: $OP^2 = OA^2 + PA^2 = r^2 + 3r^2 = 4r^2$, so $OP = 2r$. ✓

Example: For a circle of radius 5 cm, the tangents are $5\sqrt{3} \approx 8.66$ cm long and OP = 10 cm.

Result 3: triangle PAB is equilateral

Step 1: PA = PB (tangents from an external point), so triangle PAB is isosceles and ∠PAB = ∠PBA.

Step 2: Angle sum in triangle PAB:

$60^\circ + 2\angle PAB = 180^\circ$

$\angle PAB = 60^\circ$

Step 3: All three angles are 60°, so triangle PAB is equilateral, and the chord of contact AB = PA = $r\sqrt{3}$.

Result 4: ∠AOB = 120°

In quadrilateral OAPB, ∠OAP + ∠OBP = 180°, so ∠AOB + ∠APB = 180°:

$\angle AOB = 180^\circ - 60^\circ = 120^\circ$

Summary table

Exam tips

• Always state why ∠OAP = 90° (radius ⊥ tangent) and why OP bisects ∠APB (congruent triangles).
• You may prove OP = 2r either with sin 30° or with Pythagoras after finding PA.
• In the equilateral-triangle proof, quote "tangents from an external point are equal" for PA = PB.
• Mark the right angles and the 30° angles on your figure.

Common mistakes

• Using 60° instead of 30° in the trigonometric ratio. OP bisects the angle, so use 30°.
• Using cos instead of sin for OA/OP. OA is opposite ∠APO.
• Forgetting to justify that OP bisects the angle between the tangents.
• Writing ∠AOB = 60°. The angles at O and P are supplementary, so ∠AOB = 120°.

Frequently Asked Questions

If two tangents are inclined at 60°, what is OP?

OP = 2r. OP bisects the 60° angle, and in the right triangle OAP, sin 30° = r/OP, so OP = 2r.

What is the length of each tangent when the angle between them is 60°?

Each tangent is r√3 long, because tan 30° = r/PA gives PA = r√3.

Why is triangle PAB equilateral?

PA = PB, so the base angles are equal. With ∠APB = 60°, each base angle is (180° − 60°)/2 = 60°, so all three angles are 60°.

What is the angle between the radii OA and OB?

∠AOB = 120°, because in quadrilateral OAPB the angles at A and B are 90° each, so ∠AOB + ∠APB = 180°.

Why does OP bisect the angle between the tangents?

Triangles OAP and OBP are congruent by RHS: OA = OB, OP is common and both have a right angle. So ∠APO = ∠BPO.