In the figure given, $O$ is the centre of a circle circumscribing a quadrilateral $A B C D$ . If $A B = B C$ and $∠ B A C = 40^{o}$ , then what is $∠ A D C$ equal to?

50^{o}

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

60^{o}

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

70^{o}

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

80^{o}

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

Open in App

Solution

The correct option is D $80^{o}$
R.E.F image
$A B = B C$
$\Rightarrow x = 40^{\circ}$
In $Δ A B C$
$x + y + 40^{\circ} = 180^{o}$
$\Rightarrow y = 100^{\circ}$
$A B C D$ is a cyclic quadrilateral
$\Rightarrow y + z = 180^{o}$
$\Rightarrow z = 80^{o}$

Suggest Corrections

Similar questions

A B C D

is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and if

∠ A D C = 140^{o}

, then

∠ B A C

is equal to:

In the given figure ABCD is quadrilateral circumscribing a circle with centre O. If $∠$ B = $90^{0}$ , AD = 23 cm, AB = 29 cm and DS = 5 cm, then the radius of the circle is ___ cm.

Q. ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and

∠ A D C = 140^{\circ},

then

∠

BAC is equal to:

In the given figure, ABCD is a quadrilateral circumscribing a circle with centre 'O'. If ∠ B = 90^{\circ}, A D = 23 c m, A B = 29 c m and D S = 5 c m, then find the radius of the circle.

Q. Which of the following is the correct match with respect to the given question ?
'ABCD' is a rhombus such that

∠ A C B = 40^{o}

then

∠ A D B =

____.

$70^{o}$	As $∠ A D B = 1 / 2 (180^{o} - 40^{o})$
$45^{o}$	$∠ C + ∠ D = 90^{o}$
$50^{o}$	$∠ B C D = 80^{o} \Rightarrow ∠ A D C = 100^{o} ∴ ∠ A D B = (1 / 2) x 100^{o}$
$60^{0}$	All sides are equal

Join BYJU'S Learning Program

In the figure given, O is the centre of a circle circumscribing a quadrilateral ABCD. If AB=BC and ∠BAC=40o, then what is ∠ADC equal to?

The correct option is D 80oR.E.F imageAB=BC⇒x=40∘In ΔABCx+y+40∘=180o⇒y=100∘ABCD is a cyclic quadrilateral ⇒y+z=180o⇒z=80o

In the figure given, $O$ is the centre of a circle circumscribing a quadrilateral $A B C D$ . If $A B = B C$ and $∠ B A C = 40^{o}$ , then what is $∠ A D C$ equal to?

The correct option is D $80^{o}$
R.E.F image
$A B = B C$
$\Rightarrow x = 40^{\circ}$
In $Δ A B C$
$x + y + 40^{\circ} = 180^{o}$
$\Rightarrow y = 100^{\circ}$
$A B C D$ is a cyclic quadrilateral
$\Rightarrow y + z = 180^{o}$
$\Rightarrow z = 80^{o}$