Explanation
We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$. The given angles are $$\angle A=90^o, \angle B=6x-5^o, \angle C=7x-15^o, \angle D=2x+5^o$$.
Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$
$$\implies$$ $$90^o+6x-5^o+7x-15^o+2x+5^o=360^o$$
$$\implies$$ $$15x+75^o=360^o$$
$$\implies$$ $$15x=360^o-75^o=285^o$$
$$\implies$$ $$x=19^o$$.
Therefore, option $$B$$ is correct.
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 largest angle $$=144^o$$.
Hence, option $$D$$ is correct.
Given the two angles are $$60^o$$ and $$40^o$$.
Also, the other two angles are in the ratio $$8:5$$.
Let the angles be $$8x,5x$$, respectively.
$$\Rightarrow 60^o+40^o+8x+5x=360^o$$
$$\Rightarrow 100^o+13x=360^o$$
$$\Rightarrow 13x=360^o-100^o$$
$$\Rightarrow 13x=260^o$$
$$\therefore x=20^o$$.
$$\therefore 8x=8 \times 20^o =160^o$$,
and $$ 5x=5\times 20^o =100^o$$.
$$\therefore$$ Tmeasure of other two angles are $$160^o$$ and $$100^o.
Hence, options $$B$$ and $$C$$ are correct.
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