Explanation
The given angles are, $$80^\circ,\ 95^\circ,\ 120^\circ$$.
Let the fourth angle be $$x^\circ$$
Angle sum property states that the sum of angles of a quadrilateral is $$360^\circ$$.
Then, $$80^\circ+95^\circ+120^\circ+x^\circ=360^\circ$$
$$\Rightarrow 360^\circ - (80^\circ + 95^\circ + 120^\circ ) =x^{\circ}$$
$$\Rightarrow$$ $$x^\circ=360^\circ - 295^\circ$$
$$ =65^{\circ}$$
Therefore, the fourth angle is $$x^\circ=65^\circ$$
Hence, option $$B$$ is correct.
Given 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 $$360^o$$.
$$\Rightarrow 3x+4x+5x+6x=360^o$$
$$\Rightarrow 18x=360^o$$
$$\therefore x=20^o$$.
$$\therefore \angle A =3x=3 \times 20^o =60^o$$,
$$ \angle B =4x=4\times 20^o =80^o$$,
$$ \angle C =5x=5 \times 20^o =100^o$$
and $$ \angle D =6x=6 \times 20^o =120^o$$.
$$\therefore$$ The greatest angle is $$=120^o$$
and the smallest angle is $$=60^o$$.
Therefore, the difference between the greatest and the smallest angle is $$=120^o-60^o=60^o$$.
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.
Given ratio of angles of quadrilateral $$ABCD$$ is $$3:7:6:4$$
Let the angles of quadrilateral $$ABCD$$ be $$3x,7x,6x,4x$$, respectively.
$$\Rightarrow 3x+7x+6x+4x=360^o$$
$$\Rightarrow 20x=360^o$$
$$\therefore x=18^o$$.
$$\therefore \angle A =3x=3 \times 18^o =54^o$$,
$$ \angle B =7x=7\times 18^o =126^o$$,
$$ \angle C =6x=6 \times 18^o =108^o$$
and $$ \angle D =4x=4 \times 18^o =72^o$$.
We know that in a trapezium we have two parallel lines which form two pairs of co-interiors angles, which are Supplementary.
Clearly $$\angle A+ \angle B=180^o$$ and $$\angle C+\angle D=180^o$$
Therefore, the given quadrilateral is a trapezium.
Hence, option $$C$$ is correct.
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$. The given angles are $$\angle P=100^o, \angle R=100^o, \angle S=75^o, \angle Q$$.
Then, $$\angle P+\angle Q+\angle R+\angle S=360^o$$
$$\implies$$ $$100^o+100^o+75^o+\angle Q=360^o$$.
$$\Rightarrow$$ $$\angle Q=360^o - (100^o + 100^o + 75^o )$$
$$\Rightarrow$$ $$\angle Q=360^o - 275^o =85^{\circ}$$.
Therefore, the unknown angle is $$\angle Q=85^o$$.
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$. The given angles are $$\angle A=130^o, \angle B=120^o, \angle C=x, \angle D=50^o$$.
Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$
$$\implies$$ $$130^o+120^o+x+50^o=360^o$$.
$$\implies$$ $$x=360^o - (130^o + 120^o + 50^o )$$
$$\implies$$ $$x=360^o - 300^o =60^{\circ}$$.
Therefore, the unknown angle is $$x=60^o$$.
Hence, option $$A$$ is correct.
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