Question

12. In given figure, O is the centre of the circle, PQ is a chord and PT is tangent to the circle at P. Find
ZTPQ
T
70°​

Answer 1

$\Large{\underbrace{\underline{\bf{Question\:}}}}:$

In figure O is the center of the circle, PQ is a chord and PT is tangent to the circle at P. If $\angle$POQ=70°, find $\angle$TPQ.

$\Large{\underbrace{\underline{\bf{Answer\:}}}}:$

We know that,

✯ The radius and tangent are perpendicular at their point of contact.

Since,

ꔬ OP = OQ

Here,

ꔬ POQ is a isosceles right triangle.

$:\implies\:\rm{\angle{OPQ}\:\cong\:\angle{OQP}\:}$

Now,

ꔬ In isosceles right triangle POQ,

➳ $\angle$POQ + $\angle$OPQ + $\angle$OQP = 180°

➳ 70° + $\angle$OPQ + $\angle$OPQ = 180°

➳ 2 × $\angle$OPQ = 180° - 70°

➳ $\angle$OPQ = $\tt{\dfrac{110}{2}}$

➳ $\angle$OPQ = $\bf\pink{55^{\circ}}$

Now,

ꔬ PT is the tangent to the circle at P.

$\red\checkmark\:\rm{OP\perp{PT}}$

[Note ➝ Tangent is perpendicular to radius.]

➵ $\angle$OPT = 90°

➵ $\angle$TPQ + $\angle$OPQ = 90°

➵ $\angle$TPQ + 55° = 90°

➵ $\angle$TPQ = 90° - 55°

➵ $\angle$TPQ = $\bf\purple{35^{\circ}}$

∴ Correct Answer ➝ (a) 35°