Question

14. In the given figure, O is the centre of a circle, PQ is
a chord and PT is the tangent at P. If ZPOQ = 70°,
then ZTPQ is equal to
[CBSE 2011]
(a) 35°
(b) 45°
(d) 70°
(c) 55°​

Answer 1

✬ Value = 35° ✬

Step-by-step explanation:

Given:

• O is the centre of circle.
• PQ is a chord.
• PT is the tangent.
• ∠POQ is 70°.

To Find:

• Value of ∠TPQ ?

Solution: In ∆POQ we have

• ∠POQ = 70°
• OP = OQ {radii of circle}
• ∠OPQ = ∠OQP {angles opposite to equal sides are also equal}

By angle sum property of ∆.

➮ ∠POQ + ∠OPQ + ∠OQP = 180°

➮ 70° + 2∠OPQ = 180°

➮ 2∠OPQ = 180 – 70

➮ ∠OPQ = 110/2 = 55°

➮ ∠OPQ = ∠OQP = 55°

As we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

➮ ∠TPO = 90°

➮ ∠TPQ + ∠OPQ = 90°

➮ ∠TPQ + 55° = 90°

➮ ∠TPQ = 90° – 55° = 35°

Hence, measure of ∠TPQ is 35°.

(Option A is correct)

Answer 2

Answer:

Given :-

• ▪️ O is the centre of a circle, PQ is a chord and PT is the tangent at P. If $\angle$POQ = 70°.

To Find :-

• ▪️ What is the value of $\angle$TPQ .

Solution :-

In ∆OPQ :

▪️ OP = OQ

$\implies$ $\angle$ OQP = $\angle$OPQ

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

$\implies$ 70° + 2$\angle$OPQ = 180°

$\implies$ 2$\angle$OPQ = 180° - 70° = 110°

$\implies$ $\angle$OPQ = $\sf\dfrac{\cancel{110}}{\cancel{2}}$

$\longrightarrow$ $\angle$OPQ = 55°

It is known that the tangent is perpendicular to the radius through the point of contact.

$\therefore$ $\angle$OPT = 90°

$\implies$ $\angle$OPQ + $\angle$TPQ = 90°

$\implies$ 55° + $\angle$TPQ = 90°

$\implies$ $\angle$TPQ = 90° - 55°

$\dashrightarrow$ $\angle$TPQ = 35°

$\therefore$ The value of $\angle$TPQ = 35°

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