MATHS M.C.Q. [1 Marks Each]

In $\triangle\text{AMB}$ and $\triangle\text{CMD},$ $AM = CM [M$ is the mid-point]
$BM = DM [M$ is the mid-point]
$\angle\text{AMB}=\angle\text{CMD}$ [vertically opposite angles]
By sas congruence criterion.
$\angle\text{AMB}\cong\angle\text{CMD}$
$\therefore \ \angle1=\angle4$ [by $CPCT]$

Let the angles of the triangle be $2x, 3x$ and $7x.$
Now, $2x + 3x + 7x = 180^\circ$ [Angle sum property of triangle]
$\Rightarrow 12x = 180^\circ$
$\Rightarrow x = 15^\circ$
$\therefore$ Largest angle $= 7x = 7 \times 15^\circ $
$= 105^\circ $
Hence, the correct answer is option $(c).$

A triangle has $3$ medians. The perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. A triangle has $3$ altitudes.

As we know, median of a triangle bisects the opposite sides.

Hence, the median is $BE$ as $AE = EC.$

Given, $AB | | CD$ and $AC$ is the transversal.
So, $\angle1=\angle3$
Also, in $\triangle\text{ABC},$ $\angle3+\angle4=\angle1+\angle2$
$[\because$ exterior angle = sum of two opposite interior angles$]$

$\angle\text{ECD}=\angle\text{ABC}=50^\circ$
$\therefore\text{x}=180^\circ-(65^\circ+50^\circ)$
$=65^\circ$

Under $ASA$ congruence criterion, two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle.

$\angle \text{CAD}=\angle \text{ABC}+\angle \text{ACB}$ [Exterior angle property]
$\Rightarrow 123^\circ = 39^\circ + \text{x}^\circ$
$\Rightarrow 84^\circ = \text{x}^\circ$
$\Rightarrow \text{x} = 84$
Hence, the correct answer is option $(a).$

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

In $\triangle \text{ABC}$
$\angle \text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\angle\text{B}=90^\circ-\angle\text{A}$
$\text{csc}^2\text{A}-\text{sec}^2\text{B}$
$\text{csc}^2\text{A}-\text{sec}^2(90^\circ-\angle\text{A})$
$(\therefore\text{sec}(90^\circ-\angle\text{A})=\text{csc A})$
$\text{csc}^2\text{A}-\text{csc}^2\text{A}$
$=0$

$x^\circ + x^\circ + x^\circ = 180^\circ$
$\Rightarrow x^\circ = 60^\circ .$

$\angle \text{A}-\angle \text{B}=33^\circ$ 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 $(d).$

As we know, sum of any two sides in a triangle is always greater than the third side.

$\text{In} \ \triangle\text{ABC},$
$\text{AB}+\text{BC}>\text{AC}$

Third angle $= 180^\circ - (90^\circ + 30^\circ )$
$= 60^\circ .$

As we know,
Measure of exterior angle = Sum of the opposite interior angles.
$\Rightarrow \ \angle\text{R}=\angle\text{P}+\angle\text{Q}$
$\Rightarrow \ 120^{0}=\text{x}+50^{0}$ $[\because \angle\text{R}=120^{0}]$
$\Rightarrow \ \text{x}=120^{0}-50^{0}$
$\Rightarrow \ \text{x}=70^{0}$
Now, the sum of the interior angles of a triangles is 180°.
$\therefore x + y + 50^\circ = 180^\circ $
$\Rightarrow 70^\circ + y + 50^\circ = 180^\circ $
$\Rightarrow 120^\circ + y = 180^\circ $
$\Rightarrow y = 180^\circ - 120^\circ $
$\Rightarrow y = 60^\circ $

$\angle \text{AEC}+\angle \text{AEB}=180^\circ$ [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 $(b).$

A scalene angled triangle has no sides or angles that are the same. For
Eg: The sides of a triangle are in the ratio $6 : 8 : 10$
So triangle with different sides or angles is called Scalene Angled Triangle.

$(2\text{x}-5)^\circ,\Big(3\text{x}-\frac{1}{2}\Big)+\Big(30-\frac{\text{x}}{2}\Big)^\circ=180^\circ$ [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 $\triangle \text{CDE}$
$\angle \text{CDE}+\angle \text{CED}+\angle \text{ECD}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 53^\circ+53^\circ+\angle \text{ECD}=180^\circ$
$\Rightarrow \angle \text{ECD}=74^\circ$
Since, $AB \| CD$
$\therefore \angle \text{ECD}=\angle \text{CGB}=74^\circ$ [Corresponding angles]
Now, $\angle \text{CGB}+\angle \text{BGF}=180^\circ$ [Linear pair angles]
$\Rightarrow 74^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow \text{x}=106$
Now, In $\triangle \text{EGB},$
$\angle \text{EGB}+\angle \text{BEG}+\angle \text{EBG}=180^\circ$ [Angle sum property of triangle]
$\Rightarrow 74^\circ+53^\circ+\angle \text{EBG}=180^\circ$
$\Rightarrow \angle \text{EBG}=53^\circ$
Now, $\angle \text{EBG}+\text{Reflex }\angle \text{EBG}=360^\circ$ [Complete angle]
$\Rightarrow 53^\circ+\text{y}^\circ=360^\circ$
$\Rightarrow \text{y}=307$
Hence, the correct answer is option $(a).$