Question

In the figure,AOB is a diameter and AB D is a cyclic quadrilateral . If angle ADC =120° then angle CAB is​

Answer 1

As ABCD is cyclic Qua. given,

so, AngleD + AngleB=180

120+AngleB=180

AngleB=180°-120°

therefore ABC=60°

construction BD and AC

therefore angle CDB=angleCAD as angles lies on same segments.

1/2angleCBD=1/2angleCAD

1/2×120=angleCAD

60°=CAD(answer)☜☆☞☜☆☞☜☆☞☜☆☞

Answer 2

Given: AB is diameter and ABCD is a cyclic quadrilateral and ∠ADC= 120°

To find:  ∠CAB

Step-by-step explanation:

Construction : Join A to C

Therefore we get ΔABC

Now as we know The sum of opposite angles of a cyclic quadrilateral is 180°

Therefore

∠ADC+∠ABC= 180°

120°+∠ABC= 180°

∠ABC=180°-120°= 60°

Now as AB is diameter ∴ angle forming on an arc of a semicircle is 90°

we have ∠ACB= 90°

In Δ ABC

The sum of angles of triangle is 180°

∴ ∠ABC+∠BCA+∠CAB=180°

60°+90°+∠CAB=180°

150°+∠CAB=180°

∠CAB=180°-150°

∠CAB=30°

Hence the value of ∠CAB is 30°