Maths MCQ

We have,
$2\angle \text{A}=3\angle \text{B}=6\angle \text{C}$
$\therefore 3\angle \text{B}=2\angle \text{A}$ and $6\angle \text{C}=2\angle \text{A}$
$\Rightarrow \angle \text{B}=\frac{2}{3}\angle \text{A}$ and $\angle \text{C}=\frac{2}{6}\angle \text{A}=\frac{1}{3}\angle \text{A}$
Now, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$ [Angle sum property of triangle$]$
$\Rightarrow \angle \text{A}+\frac{2}{3}\angle \text{A}+\frac{1}{3}\angle \text{A}=180^\circ$
$\Rightarrow 3\angle \text{A}+2\angle \text{A}+\angle \text{A}=180^\circ\times3$
$\Rightarrow 6\angle \text{A}=540^\circ$
$\Rightarrow \angle \text{A}=90^\circ$
$\therefore$ Smallest angle $=\angle \text{C}=\frac{1}{3}\angle \text{A}=\frac{1}{3}\times90^\circ$
$=30^\circ$
Hence, the correct answer is option $(d).$

In $\triangle \text{DEF}$
$\angle \text{DEF}+\angle \text{DFE}+\angle \text{EDF}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 110^\circ+40^\circ+\angle \text{EDF}=180^\circ$
$\Rightarrow \angle \text{EDF}=30^\circ$
Now, $\angle \text{EDF}+\angle \text{FDA}=180^\circ [$Linear pair angles$]$
$\Rightarrow 30^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=150$
Now, $\angle \text{EDF}=\angle \text{ADB}=30^\circ [$Vertically opposite angles$]$
Now, In $\triangle \text{ABD},$
$\angle \text{ADB}+\angle \text{DAB}+\angle \text{ABD}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 30^\circ+90^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
Now, $\angle \text{ABD}+\angle \text{DBC}=180^\circ [$Linear pair angles$]$
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Hence, the correct answer is option $(c).$

$\angle \text{AFE}+\angle \text{EFG}=180^\circ [$Linear pair angles$]$
$\Rightarrow 124^\circ+\angle\text{EFG}=180^\circ$
$\Rightarrow \angle\text{EFG}=56^\circ$
Since, $AB \| CD$
$\therefore \angle \text{EFG}=\angle \text{FHK}=56^\circ [$Corresponding angles$]$
$\Rightarrow \text{x}=56$
Now, $\angle \text{QKH}+\angle \text{GKH}=180^\circ [$Linear pair angles$]$
$\Rightarrow 103^\circ+\angle \text{GKH}=180^\circ$
$\Rightarrow \angle \text{GKH}=77^\circ$
Since, $AB \| CD$
$\therefore \angle \text{EGF}=\angle \text{GKH}=77^\circ [$Corresponding angles$]$
$\Rightarrow \text{y}=77$
In $\triangle \text{EHK},$
$\angle \text{EHK}+\angle \text{EKH}+\angle \text{HEK}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 56^\circ+77^\circ+\text{z}^\circ=180^\circ$
$\Rightarrow \text{z}=47$
Hence, the correct answer is option $(d).$

$\angle \text{ACB}+\angle \text{ACD}=180^\circ$
$\Rightarrow 40^\circ + \text{y}^\circ = 180^\circ$
$\Rightarrow \text{y}^\circ = 140^\circ$
$\Rightarrow \text{y} = 140$
Now, $\angle \text{ACD}=\angle \text{ABC}+\angle \text{BAC} [$Exterior angle property$]$
$\Rightarrow 3\text{x}^\circ + 4\text{x}^\circ = \text{y}^\circ$
$⇒ 7\text{x}^\circ = 140^\circ$
$\Rightarrow \text{x} = 20$
Hence, the correct answer is option $(c).$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ [$Angle sum property of triangle$]$
$ \Rightarrow 50^\circ+72^\circ+\angle \text{C}=180^\circ$
$\Rightarrow \angle \text{C}+122^\circ=180^\circ$
$\Rightarrow \angle \text{C}=58^\circ$
Now, $\text{x}^\circ=\angle\text{C} [$Vertically opposite angles$]$
$\Rightarrow \text{x}^\circ= 58^\circ$
$\Rightarrow \text{x}=58$
Hence, the correct answer is option $(c).$

$\angle \text{A}-\angle \text{B}=33^\circ$ and $\angle \text{B}-\angle \text{C}=18^\circ$
$\Rightarrow \angle \text{A}=\angle \text{B}+33^\circ$ and $\angle \text{C}=\angle \text{B}-18^\circ$
Now, $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow \angle \text{B}+33^\circ+\angle \text{B}+\angle \text{B}-18^\circ=180^\circ$
$\Rightarrow 3\angle \text{B}+15^\circ=180^\circ$
$\Rightarrow 3\angle \text{B}=165^\circ$
$\Rightarrow \angle \text{B}=55^\circ$
Hence, the correct answer is option $(d).$

$\angle \text{AEC}+\angle \text{AEB}=180^\circ [$Linear angles$]$
$\Rightarrow 79^\circ+\angle \text{AEB}=180^\circ$
$\Rightarrow \angle \text{AEB}=101^\circ$
Since, $AB \| CD$
$\angle \text{ABE}=\angle \text{ECD}=58^\circ [$Alternate angles$]$
Now, In $\triangle \text{AEB}$
$\angle \text{AEB}+\angle \text{EAB}+\angle \text{ABE}=180^\circ $[Angle sum property of triangle$]$
$\Rightarrow 101^\circ+58^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}^\circ=21^\circ$
$\Rightarrow \text{x}=21$
Now, In $\triangle \text{AEB}$
$\angle \text{AEC}+\angle \text{CAE}+\angle \text{CEA}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 79^\circ+\text{y}^\circ+3\text{x}^\circ=180^\circ$
$\Rightarrow 79^\circ+\text{y}^\circ+3(21)^\circ=180^\circ$
$\Rightarrow 79^\circ+\text{y}^\circ+63^\circ=180^\circ$
$\Rightarrow \text{y}^\circ=38^\circ$
$\Rightarrow \text{y}=38$
Hence, the correct answer is option $(b).$

$(2\text{x}-5)^\circ,\Big(3\text{x}-\frac{1}{2}\Big)+\Big(30-\frac{\text{x}}{2}\Big)^\circ=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 2\text{x}-5+3\text{x}-\frac{1}{2}+30-\frac{\text{x}}{2}=180$
$\Rightarrow 2\text{x}+3\text{x}-\frac{\text{x}}{2}-5-\frac{1}{2}+30=180$
$\Rightarrow 4\text{x}+6\text{x}-\text{x}-10-1+60=180\times2$
$\Rightarrow 9\text{x}+49=360$
$\Rightarrow \text{x}=\frac{311}{9}$
Hence, the correct answer is option $(a).$

In $\triangle \text{ABD}$
$\angle \text{ADB}+\angle \text{BAD}+\angle \text{ABD}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 61^\circ+59^\circ+\angle \text{ABD}=180^\circ$
$\Rightarrow \angle \text{ABD}=60^\circ$
$\angle \text{ABD}+\angle \text{DBC}=180^\circ [$Linear pair angles$]$
$\Rightarrow 60^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow \text{y}=120$
Now, $\angle \text{ADB}=\angle \text{GDE}=61^\circ [$Vertically opposite angles$]$
Now, In $\triangle \text{GDE},$
$\angle \text{GDE}+\angle \text{DGE}+\angle \text{GED}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 61^\circ+69^\circ+\angle \text{GED}=180^\circ$
$\Rightarrow \angle \text{GED}=50^\circ$
Now, $\angle \text{GED}+\angle \text{GEF}=180^\circ [$Linear pair angles$]$
$\Rightarrow 50^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=130$
Hence, the correct answer is option $(a).$

$\angle \text{TRS}+\angle \text{TRQ}=180^\circ [$Linear angles$]$
$\Rightarrow 5\text{x}^\circ+\angle \text{TRQ}=180^\circ$
$\Rightarrow \angle \text{TEQ}=180^\circ-5\text{x}^\circ$
Now, $\angle \text{QTR}+\angle \text{TRQ}=\angle \text{PQT} [$Exterior angle property of triangle$]$
$\Rightarrow 3\text{x}^\circ+180^\circ-5\text{x}^\circ=120^\circ$
$\Rightarrow 2\text{x}^\circ=60^\circ$
$\Rightarrow \text{x}=30$
Hence, the correct answer is option $(b).$

$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 150^\circ+\angle \text{C}=180^\circ$
$\Rightarrow \angle \text{C}=30^\circ$
Now, $\angle \text{B}+\angle \text{C}=75^\circ$
$\Rightarrow \angle \text{B}+30^\circ=75^\circ$
$\Rightarrow \angle \text{B}=45^\circ$
Hence, the correct answer is option $(b).$

Let the smallest angle be $x$, then the other angle be $4x.$
Now,
$x + 4x + 90^\circ = 180^\circ $
$\Rightarrow 5x = 90^\circ $
$\Rightarrow x = 18^\circ $
Thus, the measure of the angles are $18^\circ ,$ and $4(18)^\circ = 72^\circ $
Hence, the correct answer is option $(d).$

$\angle \text{ACD}+\angle \text{ACB}=180^\circ [$Linear angles$]$
$\Rightarrow 91^\circ+\angle \text{ACB}=180^\circ$
$\Rightarrow \angle \text{ACB}=89^\circ$
Since, $AB \| DE$
$\angle \text{DEC}=\angle \text{CAB}=46^\circ [$Alternate angles$]$
Now,
$\angle \text{ACB}+\angle \text{CAB}+\angle \text{ABC}=180^\circ [$Angle sum property of triangle$]$
$\Rightarrow 89^\circ + 46^\circ + \text{x}^\circ = 180^\circ$
$⇒ \text{x}^\circ = 45^\circ$
$\Rightarrow \text{x} = 45$
Hence, the correct answer is option $(d).$