Solution

As $\angle A R Q=\angle R Q D=25^{\circ}$ [alt. $\angle s$ ]

Also, $\angle R Q C=180^{\circ}-60^{\circ}=120^{\circ}$ (linear pair)

And, $\angle S R A=120^{\circ}$ (Corresponding angle)

Now,

$\angle S R Q=120^{\circ}+25^{\circ}$

$\angle S R Q=145^{\circ}$

Hence, the correct option is (C).

Solution

Let angle of triangle $ABC$ be $\angle A, \angle B$ and $\angle C$

Given that:

$\angle A=\angle B+\angle C$

We know that in any triangle $\angle A+\angle B+\angle C=180^{\circ}$

From equation (I) and (II), get:

$\angle A+\angle A=180^{\circ}$

$ \begin{aligned} 2 \angle A & =180^{\circ} \\ \angle A & =\dfrac{180^{\circ}}{2} \\ \angle A & =90^{\circ} \end{aligned} $

Hence, the triangle is a right triangle.

Therefore, the correct option is (D).

Solution

Given: An exterior angle of triangle is $150^{\circ}$.

Let each of the two interior opposite angle be $x$.

The sum of two interior opposite angle is equal to exterior angle of a triangle.

$ \begin{aligned} 105^{\circ} & =x+x \\ 2 x & =105^{\circ} \\ x & =52 \dfrac{1}2^{\circ} \end{aligned} $

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

Hence, the correct option is (B).

Solution

Let the angle of the triangle are $5 x, 3 x$ and $7 x$. As we know that sum of all angle of triangle is $180^{\circ}$. Now,

$5 x+3 x+7 x=180^{\circ}$

$ \begin{aligned} 15 x & =180^{\circ} \\ x & =\dfrac{180^{\circ}}{15} \\ x & =12^{\circ} \end{aligned} $

Hence, the angle of the triangle are:

$5 \times 12^{\circ}=60^{\circ}$

$3 \times 12^{\circ}=36^{\circ}$

$7 \times 12^{\circ}=84^{\circ}$

All the angle of this triangle is less than 90 degree.

Hence, the triangle is an acute angled triangle.

Solution

In triangle $ABC$, Let $\angle A=130^{\circ}$.

The bisector of the angle $B$ and $C$ are $OB$ and $OC$.

Let $\angle O B C=\angle O B A=x$ and $\angle O C B=\angle O C A=y$

In triangle $ABC$,

$ \begin{aligned} \angle A+\angle B+\angle C & =180^{\circ} \\ 130^{\circ}+2 x+2 y & =180^{\circ} \\ 2 x+2 y & =180^{\circ}-130^{\circ} \\ 2 x+2 y & =50^{\circ} \\ x+y & =25^{\circ} \end{aligned} $

That is $\angle O B C+\angle O C A=25^{\circ}$

Now, in triangle BOC:

$ \begin{aligned} \angle B O C & =180^{\circ}-(\angle O B C+\angle O C B) \\ & =180^{\circ}-25^{\circ} \\ & =155^{\circ} \end{aligned} $

Hence, the correct option is (D).

Solution

See the given figure in the question:

$ \begin{gathered} 40^{\circ}+4 x+3 x=180^{\circ} \text{ (Angles on the straight line) } \\ 4 x+3 x=180^{\circ}-40^{\circ} \\ 7 x=140^{\circ} \\ x=\dfrac{140^{\circ}}{7} \\ x=20^{\circ} \end{gathered} $

Hence, the correct option is (A).

Solution

See the given figure, producing OP, to intersect RQ at X.

Given: $OP || RS$ and $RX$ is a transversal.

So, $\angle R X P=\angle X R S$ (alternative angle)

$\angle R X P=130^{\circ}$ [Given: $\angle Q R S=130^{\circ}$ ]

$RQ$ is a line segment.

So, $\angle P X Q+\angle R X V=180^{\circ}$ [linear pair axiom]

$\angle P X Q=180^{\circ}-\angle R X P=180^{\circ}-130^{\circ}$

$\angle P X Q=50^{\circ}$

In triangle $PQX, \angle O P Q$ is an exterior angle,

Therefore, $\angle O P Q=\angle P X Q+\angle P Q X$ [exterior angle = sum of two opposite interior angles]

$110^{\circ}=50^{\circ}+\angle P Q X$

$\angle P Q X=110^{\circ}-50^{\circ}$

$\angle P Q R=60^{\circ}$

Hence, the correct option is (

Solution

Given, the ratio of angles of a triangle is $2: 4: 3$.

Let the angles of a triangle be $\angle A, \angle B$ and $\angle C$.

$\angle A=2 x, \angle B=4 x \angle C=3 x$,

$\angle A+\angle B+\angle C=180^{\circ}$ [sum of all the angles of a triangle is $180^{\circ}$ ]

$2 x+4 x+3 x=180^{\circ}$

$9 x=180^{\circ}$

$x=180^{\circ} / 9$

$=20^{\circ}$

$\angle A=2 x=2 \times 20^{\circ}=40^{\circ}$

$\angle B=4 x=4 x 20^{\circ}=80^{\circ}$

$\angle C=3 x=3 \times 20^{\circ}=60^{\circ}$

So, the smallest angle of a triangle is $40^{\circ}$.

Hence, the correct option is (B).

Solution

See the figure, $x$ and $y$ are two adjacent angles.

For $ABC$ to be a straight line, the sum of two adjacent angle must be $180^{\circ}$.

Solution

We know that in a triangle, sum of all the angles is always $180^{\circ}$. So, a triangle can’t have all angles less than $60^{\circ}$.

Solution

If an angle whose measure is more than $90^{\circ}$ but less than $180^{\circ}$ is called an obtuse angle.

We know that a triangle can’t have two obtuse angle because the sum of all the angles of it can’t be more than $180^{\circ}$. It is always equal to $180^{\circ}$.

Solution

We know that sum of all the angles in a triangle is $180^{\circ}$.

The sum of all the angles is $45^{\circ}+64^{\circ}+72^{\circ}=181^{\circ}$. So, we can’t draw any triangle having sum of all the angle $181^{\circ}$.

Solution

We know that sum of all the angles in a triangle is $180^{\circ}$.

Sum of these angles $=53^{\circ}+64^{\circ}+63^{\circ}=180^{\circ}$. So, we can draw infinitely many triangles having its angles as $53^{\circ}, 64^{\circ}$ and $63^{\circ}$.

Solution

See the given figure, $l \text{ and } m$ if a transversal intersects two parallel lines, then sum of interior angles on the same side of a transversal is supplementary.

$x+44^{\circ}=180^{\circ}$

$ \begin{aligned} & x=180^{\circ}-44^{\circ} \\ & x=136^{\circ} \end{aligned} $

Solution

No, because if it will be a right angle only when they form a linear pair.

Solution

If two intersecting line are formed right then by using linear pair axiom aniom, other three angles will be a right angle.

Solution

In the first figure, sum of two interior angle is: $132^{\circ}+48^{\circ}=180^{\circ}$ [Equal to $180^{\circ}$ ]

Hence, we know that, if sum of two interior angle are equal on the same side of $n$ is $180^{\circ}$, then they are the parallel lines.

In the second figure, sum of two interior angle is:

$73^{\circ}+106^{\circ}=179^{\circ} \neq 180^{\circ}$.

Hence, we know that, if sum of two interior angle are equal on the same side of $r$ is not equal to $180^{\circ}$, then they are not the parallel lines.

Solution

If two lines $l$ and $m$ are perpendicular to the same line $n$, then each of the two corresponding angles formed by these lines $l$ and $m$ with the line $n$ are equal to $90^{\circ}$.

Hence the line $l$ and $m$ are not perpendicular but parallel.

Solution

Given:

$OD$ is the bisector of $\angle AOC, OE$ is the bisector of $\angle BOC$ and $OD \perp OE$

To prove that point $A, O$ and $B$ are collinear that is $AOB$ are straight line.

$\angle A O C=2 \angle D O C$

$\angle C O B=2 \angle C O E$

Now, adding equations (I) and (II), get:

$\angle A O C+\angle C O B=2 \angle D O C+\angle C O E$

$\angle A O C+\angle C O B=2(\angle D O C+\angle C O E)$

$\angle A O C+\angle C O B=2 \angle D O C$

$\angle A O C+\angle C O B=2 \times 90^{\circ}$

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

$\angle A O C=180^{\circ}$

So, $\angle A O C+\angle C O B$ are forming linear pair or we can say that $AOB$ is a straight line.

Hence, point $A, O$ and $B$ are collinear.

Solution

See the given figure,

$\angle 5+\angle 6=180^{\circ}$ (Linear pair angle)

$\angle 5+120^{\circ}=180^{\circ}$

$\angle 5=180^{\circ}-120^{\circ}$

$\angle 5=60^{\circ}$

Then, $\angle 1=\angle 5 \quad[Each=60^{\circ}]$

Since, these are corresponding angles.

Hence, the line $m$ and $n$ are parallel.

Solution

According to the question,

Line $l || m$ and $t$ is the transversal.

$\angle M A B=\angle S B A[$ Alt. $\angle s]$

$\dfrac{1}{2} \angle M A B=\dfrac{1}{2} \angle S B A$

$\angle P A B=\angle Q B A$

But, $\angle P A B$ and $\angle Q B A$ are alternate angles.

Hence, $AP || BQ$.

Solution

See the given figure, $AP || BQ, AP$ and $BQ$ are the bisectors of alternate interior angles $\angle C A B$ and $\angle A B F$.

To show that $l || m$.

Now, prove that $AP || BQ$ are $t$ is transversal, therefore:

$\angle P A B=\angle A B Q$ [Alternate interior angle]

$2 \angle P A B=2 \angle A B Q$ [Multiplying both sides by 2 in equation (I)]

Since, alternate interior angle are equal.

So, if two alternate interior angle are equal then lines are parallel.

Hence, $l||m$.

Solution

According to the question:

Given:

Producing $DE$ to intersect $BC$ at $P$.

$EF || BC$ and $DP$ is the transversal,

$\angle D E F=\angle D P C\quad$… (l) [Corresponding $\angle s$ ]

See the above figure, $AB || DP$ and $BC$ is the transversal,

$\angle D P C=\angle A B C\quad$… (II) [Corresponding $\angle s$ ]

Now, from equation (I) and (II), get:

$\angle A B C=\angle D E F$

Hence, proved.

Solution

See in the figure, $BA || ED$ and $BC || EF$.

Show that $\angle ABC+\angle DEF=180^{\circ}$.

Produce a ray PE opposite to ray EF.

Prove: $BC|| EF$

Now, $\angle E P B+\angle P B C=180^{\circ} \quad$ [sum of co interior is $180^{\circ}$ ]

Now, $AB || ED$ and $PE$ is transversal line,

$\angle E P B=\angle D E F$ [Corresponding angles]

Now, from equation (I) and (II),

$\angle D E F+\angle P B C=180^{\circ}$

$\angle A B C+\angle D E F=180^{\circ}[$ Because $\angle P B C=\angle A B C]$

Hence, proved.

Solution

See in the given figure, $DE || QR$ and the line $n$ is the transversal line. $\angle E A B+\angle R B A=180^{\circ} \ldots$ (I) [The interior angles on the same side of transversal are supplementary.]

Now, $\angle P A B+\angle P B A=90^{\circ}$

Then, from triangle APB, given: $\angle A P B=180^{\circ}-(\angle P A B+\angle P B A)$

So, $\angle A P B=180^{\circ}-90^{\circ}=90^{\circ}$

Solution

Given in the question, ratio of angles is: $2: 3: 4$.

Let the angles of the triangle be $2 x, 3 x$ and $4 x$.

So,

$2 x+3 x+4 x=180^{\circ}$ [sum of angles of triangle is $180^{\circ}$ ]

$9 x=180^{\circ}$

$x=\dfrac{180^{\circ}}{9}$

$x=20^{\circ}$

Therefore, $2 x=2 \times 20^{\circ}=40^{\circ}$

$3 x=2 \times 20^{\circ}=60^{\circ}$

And, $4 x=4 \times 20^{\circ}=80^{\circ}$

Hence, the angle of the triangles are $40^{\circ}, 60^{\circ}$ and $80^{\circ}$.

Solution

Given:

In triangle $ABC$,

$\angle A=90^{\circ}$ and $A L \perp B C$

To prove: $\angle B A L=\angle A C B$

Proof: Let $\angle A B C=x$

$\angle B A L=90^{\circ}-x$

As, $\angle A=x$

$\angle C A L=x$

$\angle A B C=\angle C A L$

$\angle A B C=\angle A C B$

Hence, proved.

Solution

According to the question:

Two line $p$ and $n$ are respectively perpendicular to two parallel line $l$ and $m$, that is $P \perp l$ and $n \perp m$.

To prove that $p$ is parallel to $n$.

Given: $n \perp m$

So, $\angle 1=90^{\circ}$

Now, $P \perp l$

So, $\angle 2=90^{\circ}$

Since, 1 is parallel to $m$. So,

$\angle 2=\angle 3$

[Corresponding $\angle s$ ]

So,

$\angle 2=90^{\circ}$

From equation (I) and (II), get:

$\angle 1=\angle 3$

[each $90^{\circ}$ ]

But these are corresponding angles.

Hence, $p || n$.

Solution

Two lines $AB$ and $CD$ intersect at point $O$.

To prove: (i) $\angle A O C=\angle B O D$

(ii) $\angle A O D=\angle B O C$

Proof: (i)

Ray on stands on line CD. So,

$\angle A O C+\angle A O D=180^{\circ} \ldots$ (I) [linear pair axiom]

Similarly, ray OD stands on line $AB$. So,

$\angle A O D+\angle B O D=180^{\circ}$

Now, from equation (I) and (II), get:

$\angle A O C+\angle A O D=\angle A O D+\angle B O D$

$ \angle A O C=\angle B O D $

Hence, proved.

(ii) Ray OD stands on line $AB$. $\angle A O D+\angle B O D=180^{\circ}$ … (III) [Linear pair axiom]

Similarly, ray $OB$ stands on line $CD$. So,

$\angle D O B+\angle B O C=180^{\circ}$

From equations (III) and (IV), get:

$\angle A O D+\angle B O D=\angle D O B+\angle B O C$

$ \angle A O D=\angle B O C $

Hence, proved.

Solution

Given: in triangle $ABC$, produce $BC$ to $D$ and the bisectors of $\angle A B C$ and $\angle A C D$ meet at point T.

To prove that $\angle B T C=\dfrac{1}{2} \angle B A C$

Proof: In triangle $ABC, \angle A C D$ is an exterior angle.

$\angle A C D=\angle A B C+\angle C A B$ [We know that exterior angle of a triangle is equal to the sum of two opposite angle]

$\dfrac{1}{2} \angle A C D=\dfrac{1}{2} \angle C A B+\dfrac{1}{2} \angle A B C$ [Dividing both sides by 2 in the above equation]

$\angle T C D=\dfrac{1}{2} \angle C A B+\dfrac{1}{2} \angle A B C$

[Since, CT is the bisector of

$\angle A C D$ that is $.\dfrac{1}{2} \angle A C D=\angle T C D]$

Now, in triangle BTC,

$\angle T C D=\angle B T C+\angle C B T$ [We know that exterior angle of the triangle is equal to the sum of two opposite angles]

$\angle T C D=\angle B T C+\dfrac{1}{2} \angle A B C$

…(II) [Since, BT is the bisector of triangle

$.ABC \angle C B T=\dfrac{1}{2} \angle A B C]$

Now, from equation (I) and (II), get:

$ \begin{aligned} \dfrac{1}{2} \angle C A B+\dfrac{1}{2} \angle A B C & =\angle B T C+\dfrac{1}{2} \angle A B C \\ \dfrac{1}{2} \angle C A B & =\angle B T C \\ \dfrac{1}{2} \angle B A C & =\angle B T C \end{aligned} $

Hence, proved.

Solution

Given: Lines DE $|| QR$ and the line $DE$ intersected by transversal at $A$ and the line $QR$ intersected by transversal at $B$. Also, $BP$ and $AF$ are the bisector of angle $\angle A B R$ and $\angle C A E$ respectively.

To prove: $BP || FA$

Proof: DE $|| QR$

$\angle C A E=\angle A B R$ [Corresponding angles]

$\dfrac{1}{2} \angle C A E=\dfrac{1}{2} \angle A B R$ [Dividing both side by 2 in the above equation]

$\angle C A F=\angle A B P$ [Since, bisector of angle $\angle A B R$ and $\angle C A E$ are BP and AF respectively]

Because these are the corresponding angles on transversal line $n$ and are equal.

Hence, BP $||$ FA.

Solution

Drawn a perpendicular line from the point $p$ as $PM \perp AB$. So, $\angle P M B=90^{\circ}$

Let if possible, drown another perpendicular line $PN \perp AB$. So, $\angle P M B=90^{\circ}$

Since, $\angle P M B=\angle P N B$ it will be possible when $PM$ and $PN$ coincide with each other.

Therefore, at a given point we can draw only one perpendicular to a given line.

Solution

Given:

Let lines $l$ and $m$ are two intersecting lines. Again, let $n \perp p$ to the intersecting lines meet at point $D$.

To prove that two lines $n$ and $p$ intersecting at a point.

Proof:

Let consider that line $n$ and $p$ are intersecting each other it means lines $n$ and $p$ are parallel to each other.

$n || p$

Therefore, lines $n$ and $p$ are perpendicular to $m$ and $l$ respectively.

Now, by using equation (I), $n | p$, it means that $l$ and $m$. it is a contradiction.

Since, our assumption is wrong.

Hence, line $n$ and $p$ are intersect at a point.

Solution

If triangle is an acute triangle then all the angle will be acute angle and sum of the all angle will be $180^{\circ}$.

If a triangle is a right angle triangle then one angle will be equal to $90^{\circ}$ and remaining two angle will be acute angles and sum of all the angles will be $180^{\circ}$.

Hence, a triangle must have at least two acute angles.

Solution

Given in triangle $PQR, \angle Q>\angle R$, $PA$ is the bisector of $\angle Q P R$ and $PM \perp QR$.

To prove that $\angle A P M=\dfrac{1}{2}(\angle Q-\angle R)$

Proof: PA is the bisector of $\angle Q P R$. So,

$\angle Q P A=\angle A P R$

In angle $PQM, \angle Q+\angle P M Q+\angle Q P M=180^{\circ}$

(I) [Angle sum property of a triangle]

$\angle Q+90^{\circ}+\angle Q P M=180^{\circ} \quad[\angle P M R=90^{\circ}]$

$\angle Q=90^{\circ}-\angle Q P M$

In triangle PMR, $\angle P M R+\angle R+\angle R P M=180^{\circ}$ [Angle sum property of a triangle]

$90^{\circ}+\angle R+\angle R P M=180^{\circ}[\angle P M R=90^{\circ}]$

$\angle R=180^{\circ}-90^{\circ}-\angle R P M$

$\angle R=180^{\circ}-90^{\circ}-\angle R P M$

$\angle R=90^{\circ}-\angle R P M$

Subtracting equation (III) from equation (II), get:

$$ \begin{align*} & \angle Q-\angle R=(90^{\circ}-\angle A P M)-(90^{\circ}-\angle R P M) \\ & \angle Q-\angle R=\angle R P M-\angle Q P M \\ & \angle Q-\angle R=(\angle R P A+\angle A P M)-(\angle Q P A-\angle A P M) \quad \ldots(IV) \\ & \angle Q-\angle R=\angle Q P A+\angle A P M-\angle Q P A+\angle A P M \text{ [As, } \angle R P A=\angle Q P A] \\ & \angle Q-\angle R=2 \angle A P M \\ & \angle A P M=\dfrac{1}{2}(\angle Q-\angle R) \end{align*} $$

Hence, proved.