Question

From an external point P, two tangents PA and PB are drawn to a circle with centre O. Radii OA and OB are perpendicular to the tangents and the angle \(\angle AOB = 50^\circ\). A third tangent is drawn which intersects both PA and PB. Find the angle \(\angle APB\).

Hint:

In the quadrilateral formed by two tangents from an external point and the radii to the points of contact, the angle at the center (\(\angle AOB\)) and the angle between the tangents (\(\angle APB\)) are always supplementary. They add up to \(180^\circ\). So, a quick calculation is \(\angle APB = 180^\circ - \angle AOB\).

Answer 1

Correct Answer: 130

Approach: The radius-tangent right angles plus the given central angle live inside one quadrilateral \(OAPB\); its four angles add to \(360^\circ\), which pins down \(\angle APB\) in one line. The "third tangent" is a distractor.

Step 1: Mark the right angles.
A tangent is perpendicular to the radius at the point of contact, so
\[ \angle OAP = 90^\circ,\qquad \angle OBP = 90^\circ. \]

Step 2: Use the quadrilateral angle sum.
In quadrilateral \(OAPB\):
\[ \angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ. \]

Step 3: Substitute and solve.
\[ 90^\circ + \angle APB + 90^\circ + 50^\circ = 360^\circ \]
\[ \angle APB = 360^\circ - 230^\circ = 130^\circ. \]

Final answer: \(\angle APB = 130^\circ\).

Answer 2

Approach: Use the quadrilateral formed by O, A, P, B; since two of its angles are fixed at \(90^\circ\) by the tangent-radius property, the other two must complete \(360^\circ\) between them.

Step 1: OA \(\perp\) PA and OB \(\perp\) PB (radius meets tangent at \(90^\circ\)), so \(\angle OAP = \angle OBP = 90^\circ\).

Step 2: In quadrilateral OAPB, the four interior angles sum to \(360^\circ\):
\[ \angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ. \]

Step 3: Substitute the known angles:
\[ 90^\circ + \angle APB + 90^\circ + 50^\circ = 360^\circ \implies \angle APB = 130^\circ. \]

Note: The third tangent (which cuts across PA and PB near P) does not change this result, \(\angle APB\) is fixed the moment A, O, B are fixed; the extra tangent only creates a smaller triangle tangent to the circle, useful for other properties like equal perimeters, not for this angle.

Final Answer: \(\angle APB = 130^\circ\).