Question

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Answer question no 4.

Question is in the attachment.

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- BrainlyQueen01

Answer 1

Answer:

∠EDC = 65°

Step-by-step explanation:

Given: ∠OAB = 25°

In ΔOAB,

∠OBA = 25° [isosceles Δ]

∠AOB = 180° - (25°+25°) = 130° [sum of all int. ∠s of Δ is 180°]

∠ACB = 130°/2 = 65° [∠s made at center from same chord is half at ∠ made at circumferec of same sector]

∠EBC = 180° - (90° + 65°) = 25°

∠ABC = 2 × 25° = 50°

Here, ABCD is a cyclic quadrilateral.

so, ∠ABC + ∠ADC = 180°

∠ADC = 180° - 50° = 130°

Now, ∠EDC = 130°/2 = 65°

__________________

Amrit____

Answer 2

Hey there !!

Given :- angle oab = 25°

now , oa = ob ( radii of the circle )

therefore ,

angle oab = angle oba = 25° ( isoscles triangle )

so , angle aob = 180-50° = 130° ( sum of the all angles of ∆ )

now ,

angle aob = 2(angle acb) ( The angle subtended by an arc on the centre of the circle is double the angle subtended by it at any point on the remaining part of the circle . )

=> angle acb = 1/2angle aob

=> angle acb = 130°/2 = 65°

and

given , angle aed = angle bec = 90°

now , in ∆bec , we have :-

angle bec + angle angle acb + angle ebc = 180°

=> 90°+65+angle ebc = 180°

=> ebc = 25°

now ,

abc = 2×25° = 50°

now , abcd becomes concyclic quad.

therefore ,

abc + adc = 180°

50+ adc = 180°

=> adc = 130°

therefore angle edc = 130/2 = 65 °

HOPE IT HELPS