Question

AOB is a diameter and ABCD is a cyclic quadrilateral.If ADC-120°Find BAC​

Answer 1

Answer:

Required Answer:-

We know that, the angle subtend by the diameter or semi-circle at any point of the circle is 90°.

Then:

In the above circle, AB is a diameter because O is the centre. Then, ∠ACB = 90°.

Now:

Another property of cyclic quadrilaterals says that, the opposite angles add upto 180°. That means,

∠CDB + ∠CBA = 180°

∠BCD + ∠DAB = 180°

Considering the first equation, We have ∠CDB

⇒ 120° + ∠CBA = 180°

⇒ ∠CBA = 60°

We have got two out of three angles in ∆CBA, and the third angle is ∠BAC, which we have to find. By angle sum property of triangles::

⇒ ∠ABC + ∠BCA + ∠BAC = 180°

⇒ 60° + 90° + ∠BAC = 180°

⇒ ∠BAC + 150° = 180°

⇒ ∠BAC = 30°

Therefore:

The required unknown angle ∠BAC is 30°.

Answer 2

$\: \:$

Since, ADCB is a cyclic quadrilateral.

∠ADC + ∠CBA = 180°.

[sum of opposite angles of cyclic quadrilateral is 180°]

=>∠CBA = 180° -120° = 60° [∴ ∠ADC = 120°]

In ΔACB, ∠CAB + ∠CBA + ∠ACB = 180° [by angle sum property of a triangle]

∠CAB + 60°+ 90°= 180°

[triangle formed from diameter to the circle is 90° i.e., ∠ACB = 90°)

=> ∠CAB = 180° – 150° = 30°.