Explanation
Then, $$\angle A+\angle B+\angle C+\angle D=360^{\circ}$$.
To prove this, we join $$A$$ and $$C$$, i.e. we draw the diagonal $$AC$$.
In $$\triangle ABC$$,
$$\angle CAB+\angle ABC+\angle BCA=180^{\circ}$$ [Sum of all angles of a triangle is $$180^{\circ}$$].....$$(1)$$.
In $$\triangle ACD$$,
$$\angle CAD+\angle ADC+\angle DCA=180^{\circ}$$ [Sum of all angle of a triangle is $$180^{\circ}$$]....$$(2)$$.
Adding $$(1)$$ and $$(2)$$, we get,
$$\left(\angle CAB+\angle ABC+\angle BCA \right)+ \left(\angle CAD+\angle ADC+\angle DCA \right)=180^{\circ}+180^{\circ}$$
$$\implies$$ $$\angle ABC+\angle ADC+(CAB+CAD)+(BCA+DCA)=360^{\circ}$$
$$\implies$$ $$\angle ABC+\angle ADC+\angle BAD+\angle BCD=360^{\circ}$$
$$\implies$$ $$\therefore \angle A+\angle B+\angle C+\angle D=360^{\circ}$$.
That is, the sum of all angles of a quadrilateral is $$360^o=4\times90^o$$, i.e. $$4$$ right angles.
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
Given the ratio of angles of quadrilateral, $$ABCD$$ is $$3:4:5:6$$.
Let the angles of quadrilateral $$ABCD$$ be $$3x,4x,5x,6x$$, respectively.
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^\circ$$.
$$\Rightarrow 3x+4x+5x+6x=360^\circ$$
$$\Rightarrow 18x=360^\circ$$
$$\therefore x=20^\circ$$.
$$\angle A =3x$$
$$=3 \times 20^\circ$$
$$ =60^\circ$$
$$ \angle B =4x$$
$$=4\times 20^\circ$$
$$ =80^\circ$$
$$ \angle C =5x$$
$$=5 \times 20^\circ $$
$$=100^\circ$$
$$ \angle D =6x$$
$$=6 \times 20^\circ$$
$$ =120^\circ$$
The largest angle is equal to $$120^\circ.$$ Hence, option $$B$$ is correct.
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