Maths M.C.Q

3. 145°

Solution:

Given, $\text{PQ}||\text{RS}$

$\angle\text{PQC}=\angle\text{BRC}=60^\circ$$\big[$alternate exterior angles and $\angle\text{PQC}=60^\circ$ (given)$\big]$ and $\angle\text{DQR}$

$=\angle\text{QRA}=25^\circ$ [alternate interior angles]

$$$\big[\angle\text{DQR}=25^\circ,\text{(given)}\big]$

$\angle\text{QRS}=\angle\text{QRA}+\angle\text{ARS}$

$=\angle\text{QRA}+\big(180^\circ-\angle\text{BRS}\big)$ [linear pair axiom]

$=25^\circ+180^\circ-60^\circ=205^\circ-60^\circ=145^\circ$

2. $52\frac{1}{2}^\circ$

Solution:

An exterior angle of triangle is 150°

Let each of to two interior opposites angles be x.

We know that exterior angle of a equal to the sum of two interior opposite angles.

$105^\circ=\text{x}+\text{x}\Rightarrow2\text{x}=105^\circ$

$\Rightarrow\text{x}=\frac{1}{2}\times105^\circ=52\frac{1}{2}^\circ$

So, each of equal angle angle is $52\frac{1}{2}^\circ$

Hence, (b) is the correct answer.

4. An equilateral triangle.

Solution:

Let the angles of $\Delta\text{ABC}\ \text{be}\ \angle\text{A},\angle\text{B}\ \text{and}\ \angle\text{C}$

Given that $\angle\text{A}=\angle\text{B}+\angle\text{C}.....(1)$ 

But, in any $\Delta\text{ABC},\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ....(2)$[Angles sum property of triangle]

From equation (1) and (2), we get

$\angle\text{A}+\angle\text{A}=180^\circ\Rightarrow2\angle\text{A}=180^\circ\Rightarrow\angle\text{A}\frac{180^\circ}{2}=90^\circ$

$\Rightarrow\angle\text{A}=90^\circ$

Hence, the triangle is a right triangle and option (d) is correct.

3. 60°

Solution:

In the given figure, producing OP, to interscet RQ at X.

Since, $\text{OP}||\text{RS}$ and RX is a transversal.

So, $\angle\text{RXP}=\angle\text{XRS}$

$\Rightarrow\angle\text{RXP}=130^\circ$$\big[\because\angle\text{QRS}=130^\circ(\text{given})\big]....(\text{i})$

Now, RQ is a line segment.

So,$\angle\text{PXQ}+\angle\text{RXP}=180^\circ$ [linear pair axiom]

$\Rightarrow\angle\text{PXQ}=180^\circ-\angle\text{RXP}=180^\circ-130^\circ$ [from eq. (i)]

$\Rightarrow\angle\text{PXQ}=50^\circ$

In $\Delta\text{PQX},$ $\angle\text{OPQ}$ is an exterior angle,

$\therefore\angle\text{OPQ}=\angle\text{PXQ}+\angle\text{PQX}$

[$\because\ $exterior asngle = sum of two opposite interior angles]

$110^\circ=50^\circ+\angle\text{PQX}$

$\angle\text{PQX}=110^\circ-50^\circ$

$\angle\text{PQR}=60^\circ$ $[\because\ \angle\text{PQX}=\angle\text{PQR}]$

2. 40°

Solution:

Given that: The Ratio of angles of a triangle is 2 : 4 : 3

Let the angles of the triangle be $\angle\text{A},\angle\text{B},$ and $\angle\text{C},$

$\therefore\ \angle\text{A}=2\text{x},\angle\text{B}=4\text{x}$ and $\angle\text{C}=3\text{x}$

In $\angle\text{ABC},$ $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$

[$\because\ $Sum of angles of a triangle is 180°]

$\Rightarrow2\text{x}+4\text{x}+3\text{x}=180^\circ\Rightarrow9\text{x}=180^\circ\Rightarrow\text{x}=\frac{180^\circ}{9}=20^\circ$

$\therefore\ \angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$

$\angle\text{B}=4\text{x}=4\times20^\circ=80^\circ$

And $\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$

Hence, the smallest angles of a triangle is 40° and option (b) is correct answer.

4. 155°

Solution:

$$In $\Delta\text{ABC},$ we have $\angle\text{A}=130^\circ$

OB and OC are the bisectors of the angles B and C.

Let $\angle\text{OBC}=\angle\text{OBA}=\text{x}\ \text{and}\angle\text{OCB}=\angle\text{OCA}=\text{y}$

In $\Delta\text{ABC},$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$

$\Rightarrow130^\circ+2\text{x}+2\text{y}=180^\circ$

$\Rightarrow\text{x}+\text{y}=25^\circ$

$\text{i.e}\angle\text{OBC}+\angle\text{OCB}=25^\circ$

Now, In $\Delta\text{BOC}$

$\angle\text{BOC}=180^\circ-\big(\angle\text{OBC}+\angle\text{OCB}\big)$ (Angle sum Property)

$=180^\circ-25^\circ=155^\circ$

Hence, (d) is the correct answer.