Question

In the given fig.,<CAB=80 degree,<ABC=40 degree.Then <DAB+<ABD is equal to?

Answer 1

Angle cAB+Angle CBA+AngleACB=180(ANGLE SUM PROPERTY)

80+40+ANGLE ACB=180

ANGLE ACB=180-120

ANGLE ACB = 60

Further anwer is in attached file

Answer 2

$\mathbb{ SOLUTION }$ :

Given :
$\angle{CAB}$ = 80°
$\angle{ ABC }$ = 40°

To be found,
$\angle{ DAB }$ + $\angle{ABD }$ = ?

In the figure,
triangleCAB and triangleDAB has same base AB (AB is a chord of circle)


Now,

$\angle{ ACB }$ = $\angle{ ADB }$ [ $\therefore$ angles in the segment of the circle ]

And,

$\angle{ CAB }$ + $\angle{ CBA}$ + $\angle{ ACB }$ = 180°

[ $\therefore$ sum of all three angles of a triangle is 180° ]

=> 80° + 40° + $\angle{ ACB }$ = 180°

=> 120° + $\angle{ ACB }$ = 180°

=> $\angle{ ACB }$ = 180° - 120°

=> $\angle{ ACB }$ = 60°

Now,
$\angle{ ACB }$ = $\angle{ ADB }$
so,
$\angle{ ACB }$ = 60°
$\angle{ ADB }$ = 60°

$\therefore$ $\angle{ DAB }$ + $\angle{ABD }$ = 60° + 60° = 120°