Question

If TP and TQ are two tangents to a circle with centre O from an external point T so that $\angle\text{POQ} = 120^\circ$, then $\angle\text{PTQ}$ is equal to :

• A. $60^\circ$
• B. $70^\circ$
• C. $80^\circ$
• D. $90^\circ$

Hint:

Keep in mind that the angle between two tangents drawn from an external point to a circle is always supplementary to the angle subtended by the line segments joining the points of contact at the centre.
\[ \angle\text{PTQ} = 180^\circ - \angle\text{POQ} \]
Just subtract the given angle from $180^\circ$ to find the answer instantly: $180^\circ - 120^\circ = 60^\circ$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question relates to the properties of tangents drawn to a circle from an external point.
We are given:
- A circle with centre O.
- Tangents TP and TQ drawn from an external point T, touching the circle at points P and Q respectively.
- The angle subtended by the chord PQ at the centre, i.e., $\angle\text{POQ} = 120^\circ$.
We need to find the angle between the two tangents, which is $\angle\text{PTQ}$.

Step 2: Key Formula or Approach:
A key theorem in circle geometry states that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, $OP \perp PT$ and $OQ \perp QT$.
This gives:
\[ \angle\text{OPT} = 90^\circ \]
\[ \angle\text{OQT} = 90^\circ \]
In the quadrilateral $OPTQ$, the sum of all interior angles is $360^\circ$.

Step 3: Detailed Explanation:

• Identify the quadrilateral formed by the centre, the points of contact, and the external point:
This quadrilateral is $OPTQ$.

• Apply the angle sum property of quadrilaterals to $OPTQ$:
\[ \angle\text{OPT} + \angle\text{POQ} + \angle\text{OQT} + \angle\text{PTQ} = 360^\circ \]

• Substitute the known values ($\angle\text{OPT} = 90^\circ$, $\angle\text{OQT} = 90^\circ$, and $\angle\text{POQ} = 120^\circ$):
\[ 90^\circ + 120^\circ + 90^\circ + \angle\text{PTQ} = 360^\circ \]

• Sum the known constant angles:
\[ 300^\circ + \angle\text{PTQ} = 360^\circ \]

• Solve for the unknown angle $\angle\text{PTQ}$:
\[ \angle\text{PTQ} = 360^\circ - 300^\circ \]
\[ \angle\text{PTQ} = 60^\circ \]

• Alternatively, because the two opposite radial angles sum to $180^\circ$, the angle between the tangents and the angle subtended by the points of contact at the centre are supplementary:
\[ \angle\text{PTQ} + \angle\text{POQ} = 180^\circ \]
\[ \angle\text{PTQ} + 120^\circ = 180^\circ \implies \angle\text{PTQ} = 60^\circ \]


Step 4: Final Answer:
The angle $\angle\text{PTQ}$ is equal to $60^\circ$.
Hence, option (A) is correct.