If TP and TQ are two tangents drawn to a circle with center O such that - POQ -110- then- - PTQ is equal to-A- 60-B- 70-C- 80-D- 90-

The correct option is B 70° We know that ∠ OQT = ∠ OPT= 90^∘. Also ∠ OQT + ∠ OPT + ∠ POQ + ∠ PTQ = 360^∘. ⇒∠ PTQ = 360^∘ – 90^∘ – 90^∘ – 110^∘ = 70^∘ Therefo ...

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Standard X

Mathematics

Tangents from a Given Point

If TP and TQ ...

Question

If TP and TQ are two tangents drawn to a circle with center O such that $∠$ POQ $= 110^{\circ}$ , then, $∠$ PTQ is equal to:

60°

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70°

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80°

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90°

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Solution

The correct option is B
70°

We know that $∠ O Q T = ∠ O P T = 90^{\circ}$ .

Also $∠ O Q T + ∠ O P T + ∠ P O Q + ∠ P T Q = 360^{\circ}$ .

$\Rightarrow ∠ P T Q = 360^{\circ} - 90^{\circ} - 90^{\circ} - 110^{\circ}$

$= 70^{\circ}$

Therefore, $∠ P T Q = 70^{\circ}$

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Similar questions

Q. Question 2
In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that

∠ P O Q = 110^{\circ}, t h e n ∠ P T Q

is equal to
(A)

60^{\circ}

(B)

70^{\circ}

(C)

80^{\circ}

(D)

90^{\circ}

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

(A) 60 (B) 70

If TP and TQ are two tangents to a circle with center O such that $∠ P O Q = 110^{\circ}$ , then, $∠$ PTQ is equal to:

In figure displayed below, if TP and TQ are the two tangents to a circle with centre O so that $∠ P O Q = 110^{\circ}, t h e n ∠ P T Q$ is equal to

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If TP and TQ are two tangents drawn to a circle with center O such that ∠POQ =110∘, then, ∠PTQ is equal to:

The correct option is B 70° We know that ∠OQT=∠OPT=90∘. Also ∠OQT+∠OPT+∠POQ+∠PTQ=360∘. ⇒∠PTQ=360∘–90∘–90∘–110∘ =70∘ Therefore, ∠PTQ=70∘

If TP and TQ are two tangents drawn to a circle with center O such that $∠$ POQ $= 110^{\circ}$ , then, $∠$ PTQ is equal to:

The correct option is B
70°

We know that $∠ O Q T = ∠ O P T = 90^{\circ}$ .

Also $∠ O Q T + ∠ O P T + ∠ P O Q + ∠ P T Q = 360^{\circ}$ .

$\Rightarrow ∠ P T Q = 360^{\circ} - 90^{\circ} - 90^{\circ} - 110^{\circ}$

$= 70^{\circ}$

Therefore, $∠ P T Q = 70^{\circ}$