A circle of radius 10 cm has its centre at the vertex C of an equilateral triangle ABC and passes through the other two vertices. The side AC extended through C intersects the circle at D. The number of degrees of angle ADB is:

15^{0}

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

30^{0}

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

60^{0}

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

90^{0}

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

120^{0}

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

Open in App

Solution

The correct option is B $30^{0}$
$∠ A B C = 60 ° ∴ ∠ C A B = 60 °$
$∠ A B D = 90 °$ ( $∵$ angle inside the semicircle = 90° )
$∴ ∠ A D B = 180 ° - (∠ C A B + ∠ A B D) = 30 °$

Suggest Corrections

Similar questions

Q. A rhombus,

A B C D

has sides of length

10

. A circle with centre

A

, passes through

C

(the opposite vertex). Likewise, a circle with centre

B

passes through

D

. If the two circles are tangent to each other, what is the area of rhombus?

Q. ABC is an isosceles triangle and a circle such that it passes through vertex C and AB acts as a tangent at D for the same circle. AC and BC intersects the circle at E and F respectively AC = BC = 4 cm and AB = 6 cm. Also, D is the mid-point of AB. What is the ratio of EC : (AE + AD)?

ABC is the right angled triangle with $∠$ ABC= $90^{\circ}$ , The centre of the circle passing through ABC lies on _______.

ABC is an isosceles triangle with AB = AC = 4cm . A circle through B touches side AC at its middle point D and intersects side AB in point P. Find the value of AP

Q. A circle passes through the three vertices of an equilateral triangle that has sides of length

3

. The area of the circle is -

Join BYJU'S Learning Program

A circle of radius 10 cm has its centre at the vertex C of an equilateral triangle ABC and passes through the other two vertices. The side AC extended through C intersects the circle at D. The number of degrees of angle ADB is:

The correct option is B 300∠ABC=60°∴∠CAB=60° ∠ABD=90° (∵ angle inside the semicircle = 90° )∴∠ADB=180°−(∠CAB+∠ABD)=30°

The correct option is B $30^{0}$
$∠ A B C = 60 ° ∴ ∠ C A B = 60 °$
$∠ A B D = 90 °$ ( $∵$ angle inside the semicircle = 90° )
$∴ ∠ A D B = 180 ° - (∠ C A B + ∠ A B D) = 30 °$