Maths MCQ

$\text{AOB}$ is a straight line
$x + y = 180^\circ $
But $4x = 5y$
$\therefore\frac{5}{4}\text{y}+\text{y}=180^\circ$
$\frac{\text{5y}+\text{4y}}{4}=180^\circ$
$\Rightarrow\frac{9\text{y}}{4}=180^\circ$
$\text{y}=\frac{180^\circ\times4}{9}=80^\circ$
$\therefore\text{x}=\frac{5}{4}\text{y}=\frac{5}{4}\times80=100^\circ$

In the figure, $\text{AB}\parallel\text{CD}$
$\angle\text{OAB}=150^\circ,\angle\text{OCD}=120^\circ$
From $O$, draw $\text{OE}\parallel\text{AB}$ or $CD$
$\text{AB}\parallel\text{DE}$
$\angle\text{OAB}+\angle\text{AOE}=180^\circ$
$\Rightarrow150^\circ+\angle\text{AOE}=180^\circ$
$\Rightarrow\angle\text{AOE}=180^\circ-150^\circ=30^\circ$
Similarly $\text{DE}\parallel\text{CD}$
$\Rightarrow\angle\text{EOC}+\angle\text{OCD}=180^\circ$
$\Rightarrow\angle\text{EOC}+120^\circ=180^\circ$
$\Rightarrow\angle\text{EOC}=180^\circ-120^\circ=60^\circ$
Now $\angle\text{AOC}=\angle\text{AOE}+\angle\text{EOC}$
$=30^\circ+60^\circ=90^\circ$

Given:
$2\angle\text{A}=3\angle\text{B}$
$\angle\text{A}=\frac{3}{2}\angle\text{B}\ \dots(\text{i})$
$3\angle\text{B}=6\angle\text{C}$
$=\angle\text{C}=\frac{1}{2}\angle\text{B}\ \dots(\text{ii})$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\frac{3}{2}\angle\text{B}+\angle\text{B}+\frac{1}{2}\angle\text{B}=180^\circ$
$\Rightarrow(\frac{3}{2}+1+\frac{1}{2})\angle\text{B}=180^\circ$
$\Rightarrow\frac{6}{2}\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=\frac{180^\circ}{3}=60^\circ$

In the figure,
Side $BC$ of $\triangle\text{ABC}$ is produced to $D$
$CE \|  BA$ is drawn
$\angle\text{A}=50^\circ$ and $\angle\text{ECD}=60^\circ$
$\text{AB}\|\text{CE}$
$\angle\text{ABC}=\angle\text{ECD} ($corresponding angle$)=60^\circ$
But in $\triangle\text{ABC,}$
$\angle\text{A}+\angle\text{B }+\angle\text{ACB }=180^\circ ($Angles of triangles$)$
$\Rightarrow50^\circ+60^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-50^\circ-60^\circ=70^\circ$

$\text{AOB}$ is a straight line
$\angle\text{AOC}+\angle\text{COD+}\angle\text{DOB}=180^\circ$
$\Rightarrow 3x - 8^\circ + 50^\circ + x + 10^\circ = 180^\circ $
$\Rightarrow 4x = 180^\circ + 8^\circ - 50^\circ - 10^\circ $
$\Rightarrow 4x = 128^\circ $
$\Rightarrow x = 32^\circ $

In the figure,
$\angle\text{AOB}=50^\circ,\angle\text{BOC}=90^\circ$
$\angle\text{COD}=70^\circ,\angle\text{AOD}=\text{x}^\circ.$
But $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}+\angle\text{DOA}$ (Angles at point)
$\Rightarrow50^\circ+90+^\circ+70+\text{x}=360^\circ$
$\Rightarrow210+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-210^\circ$
$\Rightarrow\text{x}=150^\circ$

In $\triangle\text{ABC,}$
$\angle\text{A}=65^\circ-\angle\text{B}$
$\angle\text{B}+\angle\text{C}=140^\circ$
$\angle\text{C}=140^\circ-\angle\text{B}$
But $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ ($Angles of a triangle$)$
$\Rightarrow65^\circ-\angle\text{B}+140^\circ-\angle\text{B}+\angle\text{B}=180^\circ$
$\Rightarrow205^\circ-\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=205^\circ-180^\circ=25^\circ$

In the figure,
$\text{AOB}$ is a straight line
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$ (Linear pair)
$\Rightarrow2\text{x}-10^\circ+3\text{x}+15^\circ=180^\circ$
$5\text{x}=180^\circ+10^\circ-15^\circ=175^\circ$
$\Rightarrow\text{x}=35^\circ$
$\text{x}=35$

Two supplementary angle are in the ratio $= 3 : 2$
Let first angle $= 3x$
Second angle $= 2x$
But $3x + 2x = 180^\circ $
$\Rightarrow 5x = 180^\circ $
$\Rightarrow x = 36^\circ $
Smaller angle $= 2x = 2 \times 36^\circ = 72^\circ $

In $\triangle\text{ABC,}$ side $BC$ is produced to $D$
$\triangle\text{ACD}=132^\circ$ and $\angle\text{A}=54^\circ$
Ext. $\angle\text{ACD}=\angle\text{A}+\angle\text{B}$
$\Rightarrow132^\circ=54^\circ+\angle\text{B}$
$\Rightarrow\angle\text{B}=132^\circ-54^\circ=78^\circ$

In the figure $\angle\text{AOC}=68^\circ$
But $\angle\text{AOC}+\angle\text{BOC}=180^\circ ($Linear pair$)$
$68^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=180^\circ-68^\circ=112^\circ$

In the given figure,
$\text{PQ}\parallel\text{RS,}$
$\angle\text{PAB}=60^\circ$ and $\angle\text{ACS}=100^\circ$
$\text{PQ}\parallel\text{RS}$
$\angle\text{ABC}=\angle\text{PAB}$ (alternate angles) $60^\circ$
But Ext. $\angle\text{ACS}=\angle\text{BAC}+\angle\text{ABC}$
$\Rightarrow100^\circ\angle\text{BAC}+60^\circ$
$\Rightarrow\angle\text{BAC}=100^\circ-60^\circ=40^\circ$

In $\triangle\text{ABC,}$ side $BC$ is produced to $D$
$\angle\text{A}=45^\circ, \angle\text{B}=55^\circ,$
Ext. $\angle\text{ACD}=\angle\text{A}+\angle\text{B}$
$45^\circ+55^\circ=100^\circ$

$\angle\text{ACD}=125^\circ$
$\angle\text{ACD}=\angle\text{CAB}+\angle\text{ABC}$
$(\because$ the exterior angles are equal to the sum of its interior opposite angles$)$
$\therefore\angle\text{ABC}=125^\circ-60^\circ=65^\circ$

$AB$ and $CD$ intersect each other at $O$ and $\angle\text{AOC}=50^\circ$
$\angle\text{BOD}=\angle\text{AOC}=50^\circ ($Vertically opposite angles$)$

$\text{In}\ \triangle\text{ABC}:$
$A + B + C = 180^\circ .....(i)$
Given, $A - B = 33^\circ $
$A = 33^\circ + B .....(ii)$
$B - C = 18^\circ $
$C = B + 18^\circ .....(iii)$
Putting the values of $A$ and $B$ in equation $(i):$
$\Rightarrow B + 33^\circ + B + B - 18^\circ = 180^\circ $
$\Rightarrow 3B = 180^\circ - 15$
$\Rightarrow\text{B}=\frac{165^\circ}{3}=55^\circ$

In the figure,
$\text{AOB}$ is a straight line
$\angle\text{AOC}+\angle\text{COD}+\angle\text{DOB}=180^\circ$
$\Rightarrow55^\circ+\text{x }+45^\circ=180^\circ$
$\Rightarrow\text{x}+100^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-100^\circ=80^\circ$

Supplement of $35^\circ = 180^\circ - 35^\circ = 145^\circ $