In the given figure- O is the centre of the circle - PBA - 45- The value of - PQB is

We have ∠ APB=90^∘ [Angle in a semicircle] In Δ APB, ∠ A+∠ B+∠ P = 180^∘ [Sum of angles of a triangle is 180^∘] ⇒ y^∘+45^∘+90^∘=180^∘ ⇒ y^∘=180^∘ 45^∘ ...

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Standard IX

Mathematics

Properties of Isosceles Triangle

In the given ...

Question

In the given figure, O is the centre of the circle $∠ P B A = 45^{\circ}$ . The value of $∠ P Q B$ is

$55^{\circ}$

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$45^{\circ}$

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$65^{\circ}$

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$75^{\circ}$

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Solution

The correct option is B
$45^{\circ}$

We have $∠ A P B = 90^{\circ}$ [Angle in a semicircle]
In $Δ A P B, ∠ A + ∠ B + ∠ P = 180^{\circ}$ [Sum of angles of a triangle is $180^{\circ}$ ]
$\Rightarrow y^{\circ} + 45^{\circ} + 90^{\circ} = 180^{\circ}$
$\Rightarrow y^{\circ} = 180^{\circ} - 45^{\circ} - 90^{\circ}$
$\Rightarrow y^{\circ} = 45^{\circ}$
Now, $∠ P A B = ∠ P Q B$ [Angles in the same segment]
Hence, $∠ P Q B = 45^{\circ}$

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In the given figure, O is the centre of the circle ∠PBA=45∘. The value of ∠PQB is

The correct option is B 45∘ We have ∠APB=90∘ [Angle in a semicircle] In ΔAPB,∠A+∠B+∠P=180∘ [Sum of angles of a triangle is 180∘] ⇒y∘+45∘+90∘=180∘ ⇒y∘=180∘−45∘−90∘ ⇒y∘=45∘ Now, ∠PAB=∠PQB [Angles in the same segment] Hence, ∠PQB=45∘

In the given figure, O is the centre of the circle $∠ P B A = 45^{\circ}$ . The value of $∠ P Q B$ is