Question

Solve this !! Ploxx!!

No Spam. ​

Answer 1

Step-by-step explanation:

Given :

• OD || BC and OC is a transversal From the figure

• ∠AOD and ∠OEC are corresponding angles ∠AOD = ∠OEC

• ∠DOC and ∠OCE are alternate angles ∠DOC = ∠OCE = 30°

To Find :

• The value of x and Y

Solution :

Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

So we get ∠DOC = 2 ∠DBC

∠DBC = ½ ∠DOC

By substituting the values :

➻ ∠DBC = 30/2

➻ ∠DBC = 15

➻ y = 15°

➻ ∠ABD = ½ ∠AOD

➻ ∠ABD = 90/2

➻ ∠ABD = 45°

➻ ∠ABE = ∠ABC

➻ 15 + 45 = 60°

Consider △ ABE Using the angle sum property ∠BAE + ∠AEB + ∠ABE = 180°

➻ x + 90° + 60° = 180°

➻ x + 150° = 180°

➻ x = 180° - 150°

➻ x = 30°

Therefore, the value of x is 30o and y is 15°

Answer 2

Solution :-

Let's take two lines in the given figure . Joining the points A,C & OB

Now here ,

♦ CO = BO [ °.° Radii of circle ]

So here ∆OBE is a isosceles triangle . Since it is a isosceles triangle the equal sides opposite angles are also equal ( ∠OBC = ∠OCB = 30° or ∠OBE = ∠OCE = 30° )

Taking ∆OBE , by angle sum property ,

→ ∠BOC + ∠OBE + ∠OCE = 180°

Substituting ∠OBE = ∠OCE [ °.° Both are equal angles ]

→ ∠BOC = 180° - [ ∠OCE + ∠OCE ]

→ ∠BOC = 180° - [ 2 ∠OCE ]

As we know that , in a circle , the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle .

.°. ∠BOC = 2∠BAC

→ 120° = 2∠BAC

→ ∠BAC = 120°/2

→ ∠BAC = 60°

And here ∠BAE = ∠CAE [°.° Line AE bisects ∠A ]

Here ∠BAE = ∠BAC/2 = 60°/2 = 30° = ∠CAE

• ∠CAE = x°

• x° = 30°

Taking ∆ABE using angle sum property

→ ∠ABE + ∠EBA + ∠AEB = 180°

→ 30° + ∠EBA + 90° = 180°

→ ∠EBA + 120° = 180°

→ ∠EBA = 180° - 120°

→ ∠EBA = 60°

As we know that , in a circle , the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle .

→ ∠AOD/2 + y = 60°

→ 90°/2 + y = 60°

→ 45° + y = 60°

→ y = 60° - 45°

→ y = 15°

Hence , x = 30° & y = 15°