In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD $⊥$ PQ and OE $⊥$ RS. If $∠ D O E = 130^{\circ},$ then $∠ P D E$ (in degrees) = ___.

65

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Solution

The correct option is A 65
Since equal chords are equidistant from the centre, we have OD = OE.

$⟹ ∠ O D E = ∠ O E D = x (say)$

$∴ x + x + 130 ° = 180^{\circ} (Angle sum property)$

$⟹ 2 x = 50 ° ⟹ x = 25^{\circ}$

i.e., $∠ O D E = 25^{\circ}$

$O D ⊥ P Q ⟹ ∠ P D O = 90^{\circ}$

$∴ ∠ P D E = ∠ P D O - ∠ O D E = 90^{\circ} - 25^{\circ} = 65^{\circ}$

Thus, $∠ P D E = 65^{\circ} .$

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In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD $⊥$ PQ and OE $⊥$ RS. If $∠ D O E = 130^{\circ},$ then $∠ P D E$ (in degrees) = ___.

In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD $⊥$ PQ and OE $⊥$ RS. If $∠ D O E = 130^{\circ},$ then $∠ P D E$ = ___.

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In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD ⊥ PQ and OE ⊥ RS. If ∠DOE=130∘, then ∠PDE (in degrees) = ___. 65

The correct option is A 65Since equal chords are equidistant from the centre, we have OD = OE. ⟹∠ODE=∠OED=x (say) ∴x+x+130°=180∘ (Angle sum property) ⟹2x=50°⟹x=25∘ i.e., ∠ODE=25∘ OD⊥PQ⟹∠PDO=90∘ ∴∠PDE=∠PDO−∠ODE=90∘−25∘=65∘ Thus, ∠PDE=65∘.

In the adjoining figure, O is the centre of the circle and PQ, RS are its equal chords, OD $⊥$ PQ and OE $⊥$ RS. If $∠ D O E = 130^{\circ},$ then $∠ P D E$ (in degrees) = ___.

65