Mathematics
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
60°
70°
80°
90°
Answer
We know that,
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
So,
OP ⊥ PT and OQ ⊥ QT
∠OPT = 90° and ∠OQT = 90°
In quadrilateral OPTQ,
⇒ ∠OPT + ∠PTQ + ∠OQT + ∠POQ = 360°
⇒ 90° + ∠PTQ + 90° + 110° = 360°
⇒ ∠PTQ + 290° = 360°
⇒ ∠PTQ = 360° - 290° = 70°.
Hence, Option 2 is the correct option.
Related Questions
If tangents PA and PB from a point P to a circle with center O are inclined to each other at angle of 80°, then ∠POA is equal to
50°
60°
70°
80°
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
7 cm
12 cm
15 cm
24.5 cm
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.