Question

In the figure ,$\angle BAD=3\angle DBA$ ,find $\angle ABC$ (in degrees)

Answer 1

$\angle BAD=3\angle DBA$

in $\triangle ABD$

$\angle DAB+\angle ABD+\angle ADB=180$

$\angle DBA+104=180$

$\angle DBA=45-26$

$=19$

$\angle BAD=3\times 19=57$

in $\angle ABC$

$\angle BAD+\angle ABC+\angle ACB=180$

$\angle ABC=180-65-57$

$={58}^o$