Ashu has constructed the bisector of AB in such a way that AO=BO and AP=BP. The value of $∠ A Q O$ is.

75^{\circ}

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{82.5}^{\circ}

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90^{\circ}

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{97.5}^{\circ}

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Solution

The correct option is C $90^{\circ}$
In the figure, AO = BO and AP = BP.
$In △ A P O and △ B P O,$
AO=BO
AP=BP
OP=OP [Common side]
$∴ △ A O P ≅ △ B O P (∵ SSS rule)$
$Hence, ∠ A O Q = ∠ B O Q [CPCT].$

$In △ A O Q and △ O Q B$
$A O = B O ∠ A O Q = ∠ B O Q O Q = O Q [Common side]$
$△ A O Q ≅ △ O Q B (∵ SAS rule)$

$∴ ∠ A Q O = ∠ B Q O ∠ A Q B = 180^{\circ}, then, ∠ A Q O must be equal to 90^{\circ} .$

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Ashu has constructed the bisector of AB in such a way that AO=BO and AP=BP. The value of ∠AQO is.

Ashu has constructed the bisector of AB in such a way that AO=BO and AP=BP. The value of $∠ A Q O$ is.