Question

Fig. 10.38
5. In Fig. 10.39. A, B, C and D are four points on a
circle. AC and BD intersect at a point E such
that 2 BEC = 130° and Z ECD = 20°. Find
BAC.​

Answer 1

From image we have :-

• A, B, C and D are four points on a circle.
• AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°.

To Find :-

• ∠BAC = ?

Solution :-

As we can see BD is a straight line.

So,

→ ∠CED + ∠BEC = 180° (Linear Pair angles.)

putting value of ∠BEC = 130° ,

→ ∠CED + 130° = 180°

→ ∠CED = 180° - 130°

→ ∠CED = 50°

also,

→ ∠ECD = 20° (given) .

Than, in ∆CED , we have,

→ ∠CED + ∠ECD + ∠CDE = 180° {Angle sum Property} .

Putting both values,

→ 50° + 20° + ∠CDE = 180°

→ 70° + ∠CDE = 180°

→ ∠CDE = 180° - 70°

→ ∠CDE = 110°

Therefore,

→ ∠BAC = ∠CDE { Angles in the same segment of a circle are equal. }

→ ∠BAC = 110°. (Ans.)

Hence, ∠BAC is equal to 110°.

Learn more :-

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3.

In the fig, AB || CD,FIND x.(Hint:Prove that AOB-COD).

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Answer 2

Given :

∠BEC = 130°

∠ECD = 20°

To find :

∠BAC

Solution :

Since AC is a straight line

∠AEB + ∠BEC = 180°

∠AEB + 130° = 180° [ linear pair ]

∠AEB = 180° - 130°

∠AEB = 50°

∠DEC = ∠AEB = 50° [ vertically opp. angles]

In ∆DEC,

∠DEC + ∠ECD + ∠BDC = 180° [ angle sum property ]

50° + 20° + ∠BDC = 180°

70° + ∠BDC = 180°

∠BDC = 180° - 70°

∠BDC = 110°

Therefore, ∠BAC = ∠BDC = 110° [ anglea in the same segment are equal ]