You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Angle Subtended by Diameter on the Circle

On the circle...

Question

On the circle with centre O, points A and B are such that OA $=$ AB. A point C is located on the tangent at B to the circle such that A and C are on the opposite sides of the lines OB and AB $=$ BC. The line segment AC intersects the circle again at F. Then the ratio $∠$ BOF $: ∠$ BOC is equal to:

1 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 : 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 : 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 : 5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 : 3$
As $O A = A B$ and we know $O A = O B$ , we have $△ O A B$ is equilateral.

Since $B C$ is a tangent, $∠ O B C = 90^{o}$ and $A B = B C$

$\Rightarrow ∠ B A C = ∠ B C A = \frac{180 - 60 - 90}{2} = 15^{o}$

$∴ ∠ B O F = 30^{o}$ and $∠ B O C = 45^{o}$

Thus, $∠ B O F : ∠ B O C = 30 : 45 = 2 : 3$

Suggest Corrections

Similar questions

On the circle with centre O, points A, B are such that OA = AB. A point C is located on the tangent at B to the circle such that A and C are on the opposite sides of the line OB and AB = BC. The line segment AC intersects the circle again at F. Then the ratio $∠$ BOF: $∠$ BOC is equal to:

Q. Let

A B

be a line segment with midpoint

C

, and

D

be the midpoint of

A C

. Let

C_{1}

be the circle with diameter

A B

, and

C_{2}

be the circle with diameter

A C

. Let

E

be a point on

C_{1}

such that

E C

is perpendicular to

A B

. Let

F

be a point on

C_{2}

such that

D F

is perpendicular to

A B

, and

E

and

F

lie on opposite sides of

A B

. Then the value of

sin ∠ F E C

B

and

C

are points on the circle

x^{2} + y^{2} = a^{2}

A point

A (b, c)

lies on that circle such that

A B = A C = d

. The equation to

B C

\dots \dots

Q. Let

A, B

are two points on the curve

y = {log}_{1 / 2} (x - \frac{1}{2}) + {log}_{2} \sqrt{4 x^{2} - 4 x + 1}

and

A

is also on the circle whose radius is

\sqrt{10}

and centre at

O (0, 0)

B

lies inside the circle such that its abscissa is integer, then

AB is a diameter and AC is chord of a circle with centre O such that $∠ B A C = 30^{\circ}$ . The tangent at C intersects AB at a point D. Prove that BC =BD.

Join BYJU'S Learning Program

On the circle with centre O, points A and B are such that OA=AB. A point C is located on the tangent at B to the circle such that A and C are on the opposite sides of the lines OB and AB=BC. The line segment AC intersects the circle again at F. Then the ratio ∠BOF:∠BOC is equal to:

The correct option is B 2:3As OA=AB and we know OA=OB, we have △OAB is equilateral.Since BC is a tangent, ∠OBC=90o and AB=BC⇒∠BAC=∠BCA=180−60−902=15o∴∠BOF=30o and ∠BOC=45oThus, ∠BOF:∠BOC=30:45=2:3