Question

arc CD subtends 56 degree at center O in the circle.ad and bc are extended to meet at e. find ∠dbc and ∠e
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Answer 1

Answer:

In cyclic quadrilateral ABCD,

∠ BCD + ∠BAD = 180°

⇒ ∠ BCD + 80° = 180°

⇒ ∠ BCD = 100°

Also,

∠ DBC + ∠ BCD + ∠ BDC = 180°

⇒40 + 100 + ∠ BDC = 180°

⇒ ∠ BDC = 40°

Since,

∠ ADB is angle in semicircle,

So, ∠ ADB = 90°

In △ ABD,

∠ ABD + ∠ ADB + ∠ BAD = 180°

⇒∠ ABD + 90° + 80° = 180°

⇒ ∠ ABD = 10°

Now,

We join CD for finding ∠ BAC.

So, we can see ∠ ACB = 90°

In ∆ ABC,

∠ BAC + ∠ ACB + ∠ CBA = 180° -----(1)

Also, we know,

∠ CBA = ∠ CBD + ∠ DBA = 40 + 10 = 50°

From equation (1), we get

∠ BAC = 180 - 50 - 90 = 180 - 140 = 40°

∠ BAC = 40°.

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Answer 2

Answer:

/_DOC = 2/_DBC [Angles subtended by the same arc, on the centre and on the circle]

/_DBC = 56/2

/_DBC = 28°

/_ADB = 90° [Angle on a semicircle]

/_EDB = 180 - /_ADB [Linear pair]

/_EDB = 90°

In Triangle EDB:

/_EDB + /_DBE + /_BED = 180° [Angle sum property]

/_E = 180 - 28 - 90

/_E = 62°

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