Explanation
Given ratio of angles of quadrilateral $$ABCD$$ is $$1:2:3:4$$
Let the angles of quadrilateral $$ABCD$$ be $$1x,2x,3x,4x$$, respectively.
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\Rightarrow 1x+2x+3x+4x=360^o$$
$$\Rightarrow 10x=360^o$$
$$\therefore x=36^o$$.
$$\therefore \angle A =1x=1 \times 36^o =36^o$$,
$$ \angle B =2x=2\times 36^o =72^o$$,
$$ \angle C =3x=3 \times 36^o =108^o$$
and $$ \angle D =4x=4\times 36^o =144^o$$.
$$\therefore$$ The smallest angle $$=36^o$$.
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 obtuse.
Let the angles be$$\angle A=90^o+a, \angle B=90^o+b, \angle C=90^o+c, \angle D=90^o+d$$.
Then, $$\angle A+\angle B+\angle C+\angle D$$
$$=90^o+a+90^o+b+90^o+c+90^o+d$$
$$=360^o +a+b+c+d >360^o.$$
Therefore, all the four angles of a quadrilateral cannot be obtuse.
That is, the statement is false.
Given ratio of angles of quadrilateral $$PQRS$$ is $$1:2:3:4$$
Let the angles of quadrilateral $$PQRS$$ be $$1x,2x,3x,4x$$, respectively.
$$\Rightarrow 1x+2x+3x+4x=360^o$$$$\Rightarrow 10x=360^o$$
$$\therefore \angle P =1x=1 \times 36^o =36^o$$,
$$ \angle Q =2x=2\times 36^o =72^o$$,
$$ \angle R =3x=3 \times 36^o =108^o$$
and $$ \angle S =4x=4\times 36^o =144^o$$.
$$\therefore$$ $$\angle S=144^o$$.
Hence, option $$D$$ is correct.
Let the four angles be $$ \angle A , \angle B , \angle C$$ and $$\angle D$$ .
Given $$ \angle A=47^o , \angle B =102^o, \angle C = 111^{o} $$ .
$$ \angle A + \angle B + \angle C + \angle D = 360^{o} $$.
Then, $$ \angle D $$ will be given by:
$$\angle D = 360^o - \angle A - \angle B-\angle C $$
$$ \Rightarrow \angle D = 360 ^o-47^o -102^o -111^o $$
$$ \Rightarrow \angle D = 360 - 260^o$$
$$ \Rightarrow \angle D = 100^o$$.
The measure of fourth angle is $$ 100 ^{o}$$.
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