In the given figure, 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°. Find ∠BAC.

Open in App

Solution

In ΔCDE,

∠CDE + ∠DCE = ∠CEB (Exterior angle)

⇒ ∠CDE + 20° = 130°

⇒ ∠CDE = 110°

However, ∠BAC = ∠CDE (Angles in the same segment of a circle)

⇒ ∠BAC = 110°

Suggest Corrections

Similar questions

Question 5
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $∠ B E C = 130^{\circ}$ and $∠ E C D = 20^{\circ}$ . Find $∠ B A C$ .

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. Find ∠BAC

Q. In figure,

A, B, C

and

D

are four points on a circle.

A C

and

B D

intersect at a point

E

such that

∠ B E C = 130^{o}

and

∠ E C D = 20^{o}

. Find

∠ B A C

In figure, $A, B, C$ and $D$ are four points on a circle. $A C$ and $B D$ intersect at a point $E$ such that $∠ B E C = 130^{\circ}$ and $∠ E C D = 20^{\circ}$ . Find $∠ B A C$ (in degrees).

Q. In Figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that

∠ B E C = 130^{o} a n d ∠ E C D = 20^{o}

. Then

∠ B A C

Join BYJU'S Learning Program

In the given figure, 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°. Find ∠BAC.

In ΔCDE, ∠CDE + ∠DCE = ∠CEB (Exterior angle) ⇒ ∠CDE + 20° = 130° ⇒ ∠CDE = 110° However, ∠BAC = ∠CDE (Angles in the same segment of a circle) ⇒ ∠BAC = 110°

In ΔCDE,

∠CDE + ∠DCE = ∠CEB (Exterior angle)

⇒ ∠CDE + 20° = 130°

⇒ ∠CDE = 110°

However, ∠BAC = ∠CDE (Angles in the same segment of a circle)

⇒ ∠BAC = 110°