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#### Question 1:

Define the following terms:
(i) Angle
(ii) Interior of an angle
(iii) Obtuse angle
(iv) Reflex angle
(v) Complementary angles
(vi) Supplementary angles

(i) Two rays OA and OB, with a common end-point O, form an angle AOB that is represented as $\angle AOB$.

(ii) The interior of an angle is the set of all points in its plane, which lie on the same side of OA as B and also on the same side of OB as A.

(iii) An angle greater than $90°$ but less than $180°$ is called an obtuse angle.

(iv) An angle greater than $180°$ but less than $360°$ is called a reflex angle.

(v) Two angles are said to be complementary if the sum of their measures is $90°$.

(vi) Two angles are said to be supplementary if the sum of their measures is $180°$.

#### Question 2:

If ∠A = 36°27'46" and ∠B = 28°43'39", find ∠A + ∠B.

$\angle A+\angle B=\left[\left(36°27\text{'}46\text{'}\text{'}\right)+\left(28°43\text{'}39\text{'}\text{'}\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
We know that

Deg Min Sec
36 27 46
+28 43 39
65 11 25

Hence, the measure of the required angle =

#### Question 3:

Find the difference between two angles measuring 36° and 24°28' 30".

We know that

Deg Min Sec
35 59 60
24 28 30
11 31 30

Hence, the measure of the required angle = .

#### Question 4:

Find the complement of each of the following angles.
(i) 58°
(ii) 16°
(iii) $\frac{2}{3}$ of a right angle
(iv) 46°30'
(v) 52°43'20"
(vi) 68°35'45"

(i) Complement of $58°=\left(90-58\right)°$
$=32°$

(ii) Complement of $16°=\left(90-16\right)°$
$=74°$

(iii)

$=30°$

(iv) We know that
Complement of

(v) We know that
Complement of

Deg Min Sec
89 59 60
52 43 20
37 16 40

Hence, the measure of the angle.

(vi) We know that
Complement of
Deg Min Sec
89 59 60
68 35 45
21 24 15

Hence, the measure of the required angle .

#### Question 5:

Find the supplement of each of the following angles.
(i) 63°
(ii) 138°
(iii) $\frac{3}{5}$ of a right angle
(iv) 75°36'
(v) 124°20'40"
(vi) 108°48'32"

(i) Supplement of $63°=\left(180-63\right)°$
$=117°$

(ii) Supplement of $138°=\left(180-138\right)°$
$=42°$

(iii)
Supplement of

(iv) We know that
Supplement of
$=104°24\text{'}$

(v) We know that
Supplement of

Deg Min Sec
179 59 60
124 20 40
55 39 20

Hence, the measure of the required angle.

(vi) We know that
Supplement of
Deg Min Sec
179 59 60
108 48 32
71 11 28

Hence, the measure of the required angle.

#### Question 6:

Find the measure of an angle which is
(i) equal to its complement
(ii) equal to its supplement

(i) Let the measure of the required angle be $x°$.
Then, in case of complementary angles:
$x+x=90°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, measure of the angle that is equal to its complement is $45°$.

(ii) Let the measure of the required angle be $x°$​.
Then, in case of supplementary angles:
$x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, measure of the angle that is equal to its supplement is $90°$.

#### Question 7:

Find the measure of an angle which is 36° more than its complement.

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
Therefore,
$x-\left(90°-x\right)=36°\phantom{\rule{0ex}{0ex}}⇒2x=126°\phantom{\rule{0ex}{0ex}}⇒x=63°$
Hence, the measure of the required angle is $63°$.

#### Question 8:

Find the measure of an angle that is 25° less than its supplement.

Let the measure of the required angle be $x$.
Then, measure of its supplement $=\left(180°-x\right)$.
Therefore,
$\left(180°-x\right)-x=25°\phantom{\rule{0ex}{0ex}}⇒2x=155°\phantom{\rule{0ex}{0ex}}⇒x=77.5°$
Hence, the measure of the required angle is $77.5°$.

#### Question 9:

Find the angle which is four times its complement.

Let the measure of the required angle be $x$.
Then, measure of its complement $=\left(90°-x\right)$.
Therefore,
$x=\left(90°-x\right)4\phantom{\rule{0ex}{0ex}}⇒x=360°-4x\phantom{\rule{0ex}{0ex}}⇒5x=360°\phantom{\rule{0ex}{0ex}}⇒x=72°$
Hence, the measure of the required angle is $72°$.

#### Question 10:

Find the angle which is five times its supplement.

Let the measure of the required angle be $x$.
Then, measure of its supplement $=\left(180°-x\right)$.
Therefore,
$x=\left(180°-x\right)5\phantom{\rule{0ex}{0ex}}⇒x=900°-5x\phantom{\rule{0ex}{0ex}}⇒6x=900°\phantom{\rule{0ex}{0ex}}⇒x=150°$
Hence, the measure of the required angle is $150°$.

#### Question 11:

Find the angle whose supplement is four times its complement.

Let the measure of the required angle be $x°$.
Then, measure of its complement $=\left(90-x\right)°$.
And, measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(180-x\right)=4\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒180-x=360-4x\phantom{\rule{0ex}{0ex}}⇒3x=180\phantom{\rule{0ex}{0ex}}⇒x=60$
Hence, the measure of the required angle is $60°$.

#### Question 12:

Find the angle whose complement is one-third of its supplement.

Let the measure of the required angle be $x°$.
Then, the measure of its complement $=\left(90-x\right)°$.
And the measure of its supplement$=\left(180-x\right)°$.
Therefore,
$\left(90-x\right)=\frac{1}{3}\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒3\left(90-x\right)=\left(180-x\right)\phantom{\rule{0ex}{0ex}}⇒270-3x=180-x\phantom{\rule{0ex}{0ex}}⇒2x=90\phantom{\rule{0ex}{0ex}}⇒x=45$
Hence, the measure of the required angle is $45°$.

#### Question 13:

Two supplementary angles are in the ratio 3 : 2. Find the angles.

Let the two angles be 3x and 2x, respectively.
Then,
$3x+2x=180\phantom{\rule{0ex}{0ex}}⇒5x=180\phantom{\rule{0ex}{0ex}}⇒x=36°$
Therefore, the two angles are

#### Question 14:

Two complementary angles are in the ratio 4 : 5. Find the angles.

Let the two angles be 4x and 5x, respectively.
Then,
$4x+5x=90\phantom{\rule{0ex}{0ex}}⇒9x=90\phantom{\rule{0ex}{0ex}}⇒x=10°$
Hence, the two angles are .

#### Question 15:

Find the measure of an angle, if seven times its complement is 10° less than three times its supplement.

Let the measure of the required angle be $x°$.
Then, the measure of its complement $=\left(90-x\right)°$.
And the measure of its supplement$=\left(180-x\right)°$.
Therefore,
$7\left(90-x\right)=3\left(180-x\right)-10\phantom{\rule{0ex}{0ex}}⇒630-7x=540-3x-10\phantom{\rule{0ex}{0ex}}⇒4x=100\phantom{\rule{0ex}{0ex}}⇒x=25$
Hence, the measure of the required angle is $25°$.

#### Question 1:

In the adjoining figure, AOB is a straight line. Find the value of x.

We know that the sum of angles in a linear pair is $180°$.
Therefore,
$\angle AOC+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒62°+x°=180°\phantom{\rule{0ex}{0ex}}⇒x°=\left(180°-62°\right)\phantom{\rule{0ex}{0ex}}⇒x=118°\phantom{\rule{0ex}{0ex}}$
Hence, the value of x is $118°$.

#### Question 2:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC and ∠BOD.

As AOB is a straight line, the sum of angles on the same side of AOB, at a point O on it, is $180°$.
Therefore,
$\angle AOC+\angle COD+\angle BOD=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x-6\right)°+55°+\left(x+20\right)°=180\phantom{\rule{0ex}{0ex}}⇒4x=111°\phantom{\rule{0ex}{0ex}}⇒x=27.5°$
Hence,
$\angle AOC=3x-6$
$=3×27.5-6\phantom{\rule{0ex}{0ex}}=76.5°$
and $\angle BOD=x+20$
$=27.5+20\phantom{\rule{0ex}{0ex}}=47.5°$

#### Question 3:

In the adjoining figure, AOB is a straight line. Find the value of x. Also, find ∠AOC, ∠COD and ∠BOD.

AOB is a straight line. Therefore,
$\angle AOC+\angle COD+\angle BOD=180°\phantom{\rule{0ex}{0ex}}⇒\left(3x+7\right)°+\left(2x-19\right)°+x°=180°\phantom{\rule{0ex}{0ex}}⇒6x=192°\phantom{\rule{0ex}{0ex}}⇒x=32°$
Therefore,

#### Question 4:

In the adjoining figure, x:y:z = 5:4:6. If XOY is a straight line, find the values of x, y and z.

Let
XOY is a straight line. Therefore,

$\angle XOP+\angle POQ+\angle YOQ=180°\phantom{\rule{0ex}{0ex}}⇒5a+4a+6a=180°\phantom{\rule{0ex}{0ex}}⇒15a=180°\phantom{\rule{0ex}{0ex}}⇒a=12°$
Therefore,

#### Question 5:

In the adjoining figure, what value of x will make AOB a straight line?

AOB will be a straight line if
$3x+20+4x-36=180°\phantom{\rule{0ex}{0ex}}⇒7x=196°\phantom{\rule{0ex}{0ex}}⇒x=28°$
Hence, x = 28 will make AOB a straight line.

#### Question 6:

Two lines AB and CD intersect at O. If ∠AOC = 50°, find ∠AOD, ∠BOD and ∠BOC.

We know that if two lines intersect then the vertically-opposite angles are equal.
Therefore, $\angle AOC=\angle BOD=50°\phantom{\rule{0ex}{0ex}}$
Let $\angle AOD=\angle BOC=x°$
Also, we know that the sum of all angles around a point is $360°$.
Therefore,
$\angle AOC+\angle AOD+\angle BOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒50+x+50+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=260°\phantom{\rule{0ex}{0ex}}⇒x=130°$
Hence, $\angle AOD=\angle BOC=130°$
Therefore, .

#### Question 7:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O, forming angles as shown. Find the values of x, y, z and t.

We know that if two lines intersect, then the vertically opposite angles are equal.
$\therefore \angle BOD=\angle AOC=90°\phantom{\rule{0ex}{0ex}}$
Hence, $t=90°$
Also,
$\angle DOF=\angle COE=50°$
Hence, $z=50°$
Since, AOB is a straight line, we have:
$\angle AOC+\angle COE+\angle BOE=180°\phantom{\rule{0ex}{0ex}}⇒90+50+y=180°\phantom{\rule{0ex}{0ex}}⇒140+y=180°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Also,
$\angle BOE=\angle AOF=40°$
Hence, $x=40°$

#### Question 8:

In the adjoining figure, three coplanar lines AB, CD and EF intersect at a point O. Find the value of x. Also, find ∠AOD, ∠COE and ∠AOE.

We know that if two lines intersect, then the vertically-opposite angles are equal.

Since, AOB is a straight line, we have:
$\angle AOE+\angle COE+\angle BOC=180°\phantom{\rule{0ex}{0ex}}⇒3x+5x+2x=180°\phantom{\rule{0ex}{0ex}}⇒10x=180°\phantom{\rule{0ex}{0ex}}⇒x=18°$

Therefore,
$\angle AOD=2×18°=36°\phantom{\rule{0ex}{0ex}}\angle COE=5×18°=90°\phantom{\rule{0ex}{0ex}}\angle AOE=3×18°=54°$

#### Question 9:

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each of these angles.

Let the two adjacent angles be 5x and 4x, respectively.
Then,
$5x+4x=180°\phantom{\rule{0ex}{0ex}}⇒9x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°$
Hence, the two angles are .

#### Question 10:

If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.

We know that if two lines intersect, then the vertically-opposite angles are equal.

And let $\angle BOC=\angle AOD=x$
Also, we know that the sum of all angles around a point is $360°$
$\therefore \angle AOC+\angle BOD+\angle AOD+\angle BOC=360°\phantom{\rule{0ex}{0ex}}⇒90°+90°+x+x=360°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, $\angle BOC=\angle AOD=90°$
$\therefore \angle AOC=\angle BOD=\angle BOC=\angle AOD=90°$
Hence, the measure of each of the remaining angles is 90o.

#### Question 11:

Two lines AB and CD intersect at a point O, such that ∠BOC + ∠AOD = 280°, as shown in the figure. Find all the four angles.

We know that if two lines intersect, then the vertically-opposite angles are equal.
Let $\angle BOC=\angle AOD=x°$
Then,
$x+x=280\phantom{\rule{0ex}{0ex}}⇒2x=280\phantom{\rule{0ex}{0ex}}⇒x=140°\phantom{\rule{0ex}{0ex}}\therefore \angle BOC=\angle AOD=140°$
Also, let $\angle AOC=\angle BOD=y°$
We know that the sum of all angles around a point is $360°$.
$\therefore \angle AOC+\angle BOC+\angle BOD+\angle AOD=360°\phantom{\rule{0ex}{0ex}}⇒y+140+y+140=360°\phantom{\rule{0ex}{0ex}}⇒2y=80°\phantom{\rule{0ex}{0ex}}⇒y=40°$
Hence, $\angle AOC=\angle BOD=40°$

#### Question 12:

In the given figure, ray OC is the bisector of ∠AOB and OD is the ray opposite OC. Show that ∠AOD = ∠BOD.

Since DOC is a straight line, we have:

Also, $\angle COB=\angle AOC$     [Since OC is a bisector of  $\angle AOB$]
$\therefore \angle AOD=\angle BOD$

#### Question 13:

In the given figure, AB is a mirror; PQ is the incident ray and QR, the reflected ray. If ∠PQR = 112°, find ∠PQA.

We know that the angle of incidence = angle of reflection.
Hence, let $\angle PQA=\angle BQR=x°$
Since, AQB is a straight line, we have:
$\therefore \angle PQA+\angle PQR+\angle BQR=180°\phantom{\rule{0ex}{0ex}}⇒x+112+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=68°\phantom{\rule{0ex}{0ex}}⇒x=34°\phantom{\rule{0ex}{0ex}}\therefore \angle PQA=34°$

#### Question 14:

If two straight lines intersect each other, then prove that the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

Let AB and CD be the two lines intersecting at a point O and let ray OE bisect $\angle AOC$. Now, draw a ray OF in the opposite direction of OE, such that EOF is a straight line.
Let .
We know that vertically-opposite angles are equal.

But, $\angle 1=\angle 2$    [Since OE bisects $\angle AOC$ ]
$\therefore \angle 4=\angle 3$
Hence, the ray opposite the bisector of one of the angles so formed bisects the vertically-opposite angle.

#### Question 15:

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Let AOB denote a straight line and let  be the supplementary angles.
Then, we have:

Let .
Then, we have:

Therefore,
$\angle COE+\angle FOC=\frac{1}{2}x+\frac{1}{2}\left(180°-x\right)$
$=\frac{1}{2}\left(x+180°-x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(180°\right)\phantom{\rule{0ex}{0ex}}=90°$

#### Question 1:

In the adjoining figure, AB || CD. The lines are cut by a transversal t at E and F, respectively. If ∠1 = 70°, find the measure of each of the remaining marked angles.

We have, $\angle 1=70°$. Then,

#### Question 2:

In the adjoining figure, AB || CD. The lines are cut by a transversal t at E and F, respectively. If ∠2 : ∠1 = 5 : 4, find the measure of each one of the marked angles.

Let . Then,

#### Question 3:

Let $AD\parallel BC$ and CD is the transversal. Then,

Also,
$AB\parallel CD$ and BC is the transversal. Then,

From (i) and (ii), we get:

$⇒\angle ADC=\angle ABC$

#### Question 4:

In each of the figures given below, AB || CD. Find the value of x in each case.

(i)

Draw $EF\parallel AB\parallel CD$.
Now, $AB\parallel EF$ and BE is the transversal.
Then,

Again, $EF\parallel CD$ and DE is the transversal.
Then,

(ii)

Draw $EO\parallel AB\parallel CD$.
Then, $\angle EOB+\angle EOD=x°$
Now, $EO\parallel AB$ and BO is the transversal.

Again, $EO\parallel CD$ and DO is the transversal.

Therefore,

(iii)

Draw $EF\parallel AB\parallel CD$.
Then, $\angle AEF+\angle CEF=x°$
Now, $EF\parallel AB$ and AE is the transversal.

Again, $EF\parallel CD$ and CE is the transversal.

Therefore,

#### Question 5:

In the given figure, AB || CD || EF. Find the value of x.

$EF\parallel CD$ and CE is the transversal.
Then,

Again, $AB\parallel CD$ and BC is the transversal.
Then,

#### Question 6:

In the given figures, AB || CD. Find the value of x.

Draw $EF\parallel AB\parallel CD$.
$EF\parallel CD$ and CE is the transversal.
Then,

Again, $EF\parallel AB$ and AE is the transversal.
Then,

#### Question 7:

In the given figure, AB || CD. Prove that ∠BAE − ∠DCE = ∠AEC.

Draw $EF\parallel AB\parallel CD$ through E.
Now, $EF\parallel AB$ and AE is the transversal.
Then,
Again, $EF\parallel CD$ and CE is the transversal.
Then,

#### Question 8:

In the given figure, AB || CD and BC || ED. Find the value of x.

$BC\parallel ED$ and CD is the transversal.
Then,

$AB\parallel CD$ and BC is the transversal.

#### Question 9:

In the given figure, AB || CD. Prove that p + qr = 180.

Draw $PFQ\parallel AB\parallel CD$.
Now, $PFQ\parallel AB$ and EF is the transversal.
Then,

Also, $PFQ\parallel CD$.

#### Question 10:

In the given figure, AB || PQ. Find the values of x and y.

Given, $AB\parallel PQ\phantom{\rule{0ex}{0ex}}$.
Let CD be the transversal cutting AB and PQ at E and F, respectively.
Then,

We know that the sum of angles of a triangle is $180°$.
$\therefore \angle GEF+\angle EGF+\angle EFG=180\phantom{\rule{0ex}{0ex}}⇒85°+x+25°=180°\phantom{\rule{0ex}{0ex}}⇒110°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=70°$
And

#### Question 11:

In the given figure, AB || CD. Find the value of x.

$AB\parallel CD$ and AC is the transversal.
Then,

And,

We know that the sum of the angles of a triangle is $180°$.
$\angle ECF+\angle CFE+\angle CEF=180°\phantom{\rule{0ex}{0ex}}⇒105°+30°+x=180°\phantom{\rule{0ex}{0ex}}⇒135°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=45°$

#### Question 12:

In the given figure, AB || CD. Find the value of x.

$AB\parallel CD$ and PQ is the transversal.
Then,

And,

Also,

We know that the sum of angles of a triangle is $180°$.
$⇒\angle QGH+\angle GHQ+\angle GQH=180°\phantom{\rule{0ex}{0ex}}⇒95°+65°+x=180°\phantom{\rule{0ex}{0ex}}⇒x=20°\phantom{\rule{0ex}{0ex}}\therefore x=20°$

#### Question 13:

In the given figure, AB || CD. Find the values of x, y and z.

We know that the sum of the angles of a triangle is $180°$.

#### Question 14:

In the given figure, AB || CD. Find the values of x, y and z.

$AB\parallel CD$ and let EF and EG be the transversals.
Now, $AB\parallel CD$  and EF is the transversal.
Then,

Also,

And,

We know that the sum of angles of a triangle is $180°$

#### Question 15:

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

In the given figure,

#### Question 16:

For what value of x will the lines l and m be parallel to each other?

For the lines l and m to be parallel
(i)

(ii)

#### Question 17:

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Let m and n be the two parallel lines perpendicular to the line l.
Since, l is perpendicular to m,
$\angle 1=90°$
Also, since ​l is perpendicular to m,
$\angle 2=90°$
Therefore, $\angle 1=\angle 2$
This implies that the corresponding angles are equal.
Therefore, the lines m and n are parallel to each other.

#### Question 1:

In ∆ABC, if ∠B = 76° and ∠C = 48°, find ∠A.

#### Question 2:

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.

Let the angles of the given triangle measure , respectively.
Then,

Hence, the measures of the angles are .

#### Question 3:

In ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.

Let $3\angle A=4\angle B=6\angle C=x°$.
Then,

Therefore,

#### Question 4:

In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.

Let .

$=\left(130-58\right)°\phantom{\rule{0ex}{0ex}}=72°$

$=\left(108-58\right)°\phantom{\rule{0ex}{0ex}}=50°$

#### Question 5:

In ∆ABC, ∠A + ∠B = 125° and ∠A + ∠C = 113°. Find ∠A, ∠B and ∠C.

Let .
Then,
$\angle A+\angle B+\angle A+\angle C=\left(125+113\right)°\phantom{\rule{0ex}{0ex}}⇒\left(\angle A+\angle B+\angle C\right)+\angle A=238°\phantom{\rule{0ex}{0ex}}⇒180°+\angle A=238°\phantom{\rule{0ex}{0ex}}⇒\angle A=58°$

$=\left(125-58\right)°\phantom{\rule{0ex}{0ex}}=67°$

$=\left(113-58\right)°\phantom{\rule{0ex}{0ex}}=55°$

#### Question 6:

In ∆PQR, if ∠P − ∠Q = 42° and ∠Q − ∠R = 21°, find ∠P, ∠Q and ∠R.

Then,

$\mathbf{\therefore }\angle P=42°+\angle Q$
$=\left(42+53\right)°\phantom{\rule{0ex}{0ex}}=95°$

$\therefore \angle R=\angle Q-21°$
$=\left(53-21\right)°\phantom{\rule{0ex}{0ex}}=32°$

#### Question 7:

The sum of two angles of a triangle is 116° and their difference is 24°. Find the measure of each angle of the triangle.

Let
Then,
$\therefore \angle A+\angle B+\angle A-\angle B=\left(116+24\right)°\phantom{\rule{0ex}{0ex}}⇒2\angle A=140°\phantom{\rule{0ex}{0ex}}⇒\angle A=\mathbf{70}\mathbf{°}$

$\therefore \angle B=116°-\angle A$
$=\left(116-70\right)°\phantom{\rule{0ex}{0ex}}=\mathbf{46}\mathbf{°}$

Also, in ∆ ABC:

#### Question 8:

Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.

Let .
Then,

Since,
$\angle A=\angle B\phantom{\rule{0ex}{0ex}}⇒\angle B=54°\phantom{\rule{0ex}{0ex}}\therefore \angle C=\angle A+18°$

$=\left(54+18\right)°\phantom{\rule{0ex}{0ex}}=72°$

#### Question 9:

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Let the smallest angle of the triangle be $\angle C$ and let .
Then,

$\therefore \angle A=2\angle C\phantom{\rule{0ex}{0ex}}$
$=2\left(30\right)°\phantom{\rule{0ex}{0ex}}=60°$
Also,

#### Question 10:

In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.

Let ABC be a triangle right-angled at B.
Then,

Hence, .

#### Question 11:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.

Let ABC be a triangle.
Then,$\angle A=\angle B+\angle C\phantom{\rule{0ex}{0ex}}$

This implies that the triangle is right-angled at A.

#### Question 12:

ABC is right-angled at A. If ALBC, prove that ∠BAL = ∠ACB.

We know that the sum of two acute angles of a right angled triangle is $90°$
From the right $∆ABL$, we have:

Also, from the right $∆ABC$, we have:

#### Question 13:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.

Let ABC be the triangle.
Let $\angle A<\angle B+\angle C$
Then,

Also, let $\angle B<\angle A+\angle C$
Then,

And let $\angle C<\angle A+\angle B$
Then,

Hence, each angle of the triangle is less than $90°$.
Therefore, the triangle is acute-angled.

#### Question 14:

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Let ABC be a triangle and let $\angle C>\angle A+\angle B$.
Then, we have:

Since one of the angles of the triangle is greater than $90°$, the triangle is obtuse-angled.

#### Question 15:

In the given figure, side BC of ∆ABC is produced to D. If ∠ACD = 128° and ∠ABC = 43°, find ∠BAC and ∠ACB.

Side BC of triangle ABC is produced to D.

Also, in triangle ABC,

#### Question 16:

In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E. If ∠ABD = 106° and ∠ACE = 118°, find the measure of each angle of the triangle.

Side BC of triangle ABC is produced to D.

Also, side BC of triangle ABC is produced to E.

And,

#### Question 17:

Calculate the value of x in each of the following figures.

(i)
Side AC of triangle ABC is produced to E.

Also,

Substituting the value of

(ii)
From $∆ABC$ we have:

Also,

(iii)

Also,

(iv)

(v)
From $∆ABC$, we have:

Also, from $∆EBD$, we have:

(vi)
From $∆ABE$, we have:

Also, From $∆CDE$, we have

#### Question 18:

Calculate the value of x in the given figure.

Join A and D to produce AD to E.
Then,

Side AD of triangle ACD is produced to E.
(Exterior angle property)
Side AD of triangle ABD is produced to E.
(Exterior angle property)

$⇒x°=\left(\angle CAD+\angle DAB\right)+30°+45°\phantom{\rule{0ex}{0ex}}⇒x°=55°+30°+45°\phantom{\rule{0ex}{0ex}}⇒x°=130°\phantom{\rule{0ex}{0ex}}⇒x=130$

#### Question 19:

In the given figure, AD divides ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.

Now, divide $72°$ in the ratio 1 : 3.

Hence, the angles are 18o and 54o

Given,
$AD=DB\phantom{\rule{0ex}{0ex}}⇒\angle DAB=\angle DBA=18°$

In $∆ABC$, we have:

#### Question 20:

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Side BC of triangle ABC is produced to D.

Side AC of triangle ABC is produced to E.

And side AB of triangle ABC is produced to F.

Hence, the sum of the exterior angles so formed is equal to four right angles.

#### Question 21:

In the adjoining figure, show that
A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.

In $∆ACE$ , we have :
[Sum of the angles of a triangle]
In $∆BDF$, we have :
[Sum of the angles of a triangle]​

$⇒\angle A+\angle B+\angle C+\angle D+\angle E+\angle F\mathbf{=}\mathbf{360}\mathbf{°}$

#### Question 22:

In ∆ABC, the angle bisectors of ∠B and ∠C meet at O. If ∠A = 70°, find ∠BOC.

Let

From $∆ABC$, we have:

Now in $∆OBC$, we have:

#### Question 23:

The sides AB and AC of ∆ABC have been produced to D and E, respectively. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, find ∠BOC.

Let

Then,

Again,

Now, in $∆OBC$, we have:

#### Question 24:

In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1 and ACCD. Find the measure of ∠ECD.

Let
From $∆ABC$, we have:

Side BC of triangle ABC is produced to E.
$\therefore \angle ACE=\angle A+\angle B\phantom{\rule{0ex}{0ex}}⇒\angle ACD+\angle ECD=\left(90+60\right)°\phantom{\rule{0ex}{0ex}}⇒90°+\angle ECD=150°\phantom{\rule{0ex}{0ex}}⇒\angle ECD=60°$

#### Question 25:

In the given figure, AMBC and AN is the bisector of ∠A. Find the measure of ∠MAN.

From $∆ABC$, we have:

Since AN is the bisector of $\angle A$, we have:
$\angle NAB=\frac{1}{2}\angle A\phantom{\rule{0ex}{0ex}}⇒\angle NAB=\frac{1}{2}\left(85\right)°\phantom{\rule{0ex}{0ex}}⇒\angle NAB=42.5°$

From the right $∆AMB$, we have:
$\angle B+\angle MAB=90°\phantom{\rule{0ex}{0ex}}⇒\angle MAB=90°-\angle B\phantom{\rule{0ex}{0ex}}⇒\angle MAB=\left(90-65\right)°\phantom{\rule{0ex}{0ex}}⇒\angle MAB=25°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \angle MAN=\angle NAB-\angle MAB\phantom{\rule{0ex}{0ex}}⇒\angle MAN=\left(42.5-25\right)°\phantom{\rule{0ex}{0ex}}⇒\angle MAN=17.5°$

#### Question 26:

State 'True' or 'False':
(i) A triangle can have two right angles.
(ii) A triangle cannot have two obtuse angles.
(iii) A triangle cannot have two acute angles.
(iv) A triangle can have each angle less than 60°.
(v) A triangle can have each angle equal to 60°.
(vi) There cannot be a triangle whose angles measure 10°, 80° and 100°.

(i) The given statement is false.

(ii) The given statement is true.

(iii) ​The given statement is false.

(iv) The given statement is false.

(v) The given statement is true.

(vi) The given statement is true.

#### Question 1:

If two angles are complements of each other, then each angle is
(a) an acute angle
(b) an obtuse angle
(c) a right angle
(d) a reflex angle

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is $90°$, then each angle is an acute angle.

#### Question 2:

An angle which measures more than 180° but less than 360° is called
(a) an acute angle
(b) an obtuse angle
(c) a straight angle
(d) a reflex angle

(d) a reflex angle

An angle which measures more than 180o but less than 360o is called a reflex angle.

#### Question 3:

The complement of 72°40' is
(a) 107°20'
(b) 27°20'
(c) 17°20'
(d) 12°40'

(c) 17°20'

We know that
The complement of

#### Question 4:

The supplement of 54°30' is
(a) 35°30'
(b) 125°30'
(c) 45°30'
(d) 65°30'

(b) 125°30'

We know that
The supplement of

#### Question 5:

The measure of an angle is five times its complement. The angle measures
(a) 25°
(b) 35°
(c) 65°
(d) 75°

(d) 75°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be $\left(90-x\right)°$.
$\therefore x=5\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒x=450-5x\phantom{\rule{0ex}{0ex}}⇒6x=450\phantom{\rule{0ex}{0ex}}⇒x=75$

#### Question 6:

Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. The measure of larger angle is
(a) 72°
(b) 54°
(c) 63°
(d) 36°

(b) 54°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be$\left(90-x\right)°$.
$\mathbf{\therefore }2x=3\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒2x=270-3x\phantom{\rule{0ex}{0ex}}⇒5x=270\phantom{\rule{0ex}{0ex}}⇒x=54$

#### Question 7:

Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC = ?
(a) 63°
(b) 117°
(c) 17°
(d) 153°

(b) 117°

We have:

#### Question 8:

In the given figure, AOB is a straight line. If ∠AOC + ∠BOD = 95°, then ∠COD = ?
(a) 95°
(b) 85°
(c) 90°
(d) 55°

(b) 85°

We have :

#### Question 9:

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 100°

(c) 80°

We have :

#### Question 10:

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?
(a) 96°
(b) 86°
(c) 76°
(d) 106°

(b) 86°

We have :

#### Question 11:

In the given figure, AOB is a straight line. If ∠AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then ∠COD = ?
(a) 80°
(b) 100°
(c) 120°
(d) 140°

(a) 80°

We have :

#### Question 12:

In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

(c) 80°

We have :

$\therefore \angle AOC={\left[3×30-10\right]}^{°}\phantom{\rule{0ex}{0ex}}⇒\angle AOC=80°$

#### Question 13:

Which of the following statements is false?
(a) Through a given point, only one straight line can be drawn.
(b) Through two given points, it is possible to draw one and only one straight line.
(c) Two straight lines can intersect at only one point.
(d) A line segment can be produced to any desired length.

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

#### Question 14:

An angle is one-fifth of its supplement. The measure of the angle is
(a) 15°
(b) 30°
(c) 75°
(d) 150°

(b) 30°

Let the measure of the required angle be $x°$
Then, the measure of its supplement will be ${\left(180-x\right)}^{°}$
$\therefore x=\frac{1}{5}\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒5x=180°-x\phantom{\rule{0ex}{0ex}}⇒6x=180°\phantom{\rule{0ex}{0ex}}⇒x=30°$

#### Question 15:

In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ?
(a) 60°
(b) 80°
(c) 48°
(d) 72°

(a) 60°

Let

Then, we have:

#### Question 16:

In the given figure, straight lines AB and CD intersect at O. If AOC = ɸ, ∠BOC = θ and θ = 3ɸ, then ɸ = ?
(a) 30°
(b) 40°
(c) 45°
(d) 60°

(c) 45°

We have :

#### Question 17:

In the given figure, straight lines AB and CD intersect at O. If AOC + ∠BOD = 130°, then ∠AOD = ?
(a) 65°
(b) 115°
(c) 110°
(d) 125°

(b) 115°

We have :

Now,

#### Question 18:

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,

#### Question 19:

In the given figure, AB || CD. If ∠OAB = 124° and ∠OCD = 136°, then ∠AOC = ?
(a) 80°
(b) 90°
(c) 100°
(d) 110°

(c) 100°

Draw $OL\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Therefore,
$\angle AOC=\angle AOL+\angle COL\phantom{\rule{0ex}{0ex}}⇒\angle AOC=\left(56+44\right)°\phantom{\rule{0ex}{0ex}}⇒\angle AOC=\mathbf{100}\mathbf{°}$

#### Question 20:

In the given figure, AB || CD and O is a point joined with B and D, as shown in the figure, such that ∠AOB = 35° and ∠CDO = 40°. What will be the measure of reflex ∠BOD ?
(a) 225°
(b) 265°
(c) 275°
(d) 285°

(d) 285°

Draw $EOF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
$EF\parallel CD$ and OD is the transversal.

Therefore,
$\text{Reflex}\angle BOD=\angle EOB+\angle EOD\phantom{\rule{0ex}{0ex}}⇒\angle BOD=\left(145+140\right)°\phantom{\rule{0ex}{0ex}}⇒\angle BOD=\mathbf{285}\mathbf{°}$

#### Question 21:

In the given figure, AB || CD. If ABO = 130° and ∠OCD = 110, then ∠BOC = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

(b) 60°

Draw $EOF\parallel AB\parallel CD$ through O.
Now,  is the transversal.

Also,  is the transversal.

#### Question 22:

In the given figure, AB || CD. If BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
(a) 70°
(b) 60°
(c) 50°
(d) 40°

(c) 50°

Draw $EOF\parallel AB\parallel CD$.
Now,  is the transversal.

Also,
is the transversal.

Now,

#### Question 23:

In the given figure, AB || CD. If AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
(a) 130°
(b) 150°
(c) 80°
(d) 100°

(a) 130°

Draw $OE\parallel AB\parallel CD$
Now,  is the transversal.

Also,
is the transversal.

#### Question 24:

In the given figure, AB || CD. If CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°

(c) 45°

is the transversal.

Side EC of triangle EFC is produced to D.
$\therefore \angle CEF+\angle EFC=\angle DCF\phantom{\rule{0ex}{0ex}}⇒\angle CEF+25°=80°\phantom{\rule{0ex}{0ex}}⇒\angle CEF=55°$

#### Question 25:

In the given figure, AB || CD. If APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
(a) 50°
(b) 60°
(c) 40°
(d) 35°

(a) 50°

is the transversal.

Side QR of traingle PQR is produced to D.
$\therefore \angle PQR+\angle QPR=\angle PRD\phantom{\rule{0ex}{0ex}}⇒70°+\angle QPR=120°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=50°$

#### Question 26:

In the given figure, x = ?
(a) α + β − γ
(b) α − β + γ
(c) α + β + γ
(d) α + γ − β

(c) α + β + γ

Join OC to produce it to D.
Let .
Then, .
Side OC of triangle BOC is produced to D.

Side OC of triangle AOC is produced to D.

#### Question 27:

If 3 A = 4 ∠B = 6 ∠C, then A : B : C = ?
(a) 3 : 4 : 6
(b) 4 : 3 : 2

(c) 2 : 3 : 4
(d) 6 : 4 : 3

(b) 4 : 3 : 2

Let
Then,

#### Question 28:

In ABC, if ∠A + ∠B = 125° and ∠A + ∠C = 113°, then ∠A = ?
(a) (62.5)°
(b) (56.5)°
(c) 58°

(d) 63°

(c) 58°

Let

#### Question 29:

In ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21°, ∠B = ?
(a) 95°
(b) 53°
(c) 32°
(d) 63°

(b) 53°

Let

#### Question 30:

In ABC, side BC is produced to D. If ABC = 40° and ∠ACD = 120°, then A = ?
(a) 60°
(b) 40°
(c) 80°
(d) 50°

(c) 80°

$\therefore \angle A+\angle B=\angle ACD\phantom{\rule{0ex}{0ex}}⇒\angle A+40°=120°\phantom{\rule{0ex}{0ex}}⇒\angle A=80°$

#### Question 31:

Side BC of ABC has been produced to D on left hand side and to E on right hand side such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
(a) 65°
(b) 75°
(c) 50°
(d) 55°

(b) 75°

We have :

Also,

#### Question 32:

In the given figure, BAC = 30°, ∠ABC = 50° and ∠CDE = 40°. Then ∠AED = ?
(a) 120°
(b) 100°
(c) 80°
(d) 110°

(a) 120°
From $∆ABC$, we have:

Now,

Side CE of triangle CED is produced to A.

#### Question 33:

In the given figure, BAC = 40°, ∠ACB = 90° and ∠BED = 100°. Then ∠BDE = ?
(a) 50°
(b) 30°
(c) 40°
(d) 25°

(b) 30°

In $∆ABC$, we have:

In $∆EBD$, we have:

#### Question 34:

In the given figure, BO and CO are the bisectors of B and ∠C, respectively. If ∠A = 50°, then ∠BOC = ?
(a) 130°
(b) 100°
(c) 115°
(d) 120°

(c) 115°

In $∆ABC$, we have:

In $∆OBC$, we have:

#### Question 35:

In the given figure, AB || CD. If EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 55°

(c) 70°

is the transversal.

In $∆ABE$, we have:

#### Question 36:

In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
(a) 20°
(b) 25°
(c) 30°
(d) 35°

(c) 30°
In $∆OAB$, we have:

In $∆OCD$, we have:

#### Question 37:

In ∆ABC,A : ∠B : ∠C = 3 : 2 : 1 and CDAC. Then, ∠ECD = ?
(a) 60°
(b) 45°
(c) 75°
(d) 30°

Let
Then,

Hence, the angles are

Side BC of triangle ABC is produced to E.
$\therefore \angle ACE=\angle A+\angle B\phantom{\rule{0ex}{0ex}}⇒\angle ACD+\angle ECD=90°+60°\phantom{\rule{0ex}{0ex}}⇒90°+\angle ECD=150°\phantom{\rule{0ex}{0ex}}⇒\angle ECD=60°$

#### Question 38:

In the given figure, AB || CD. If ABO = 45° and ∠COD = 100°, ∠CDO = ?
(a) 25°
(b) 30°
(c) 35°
(d) 45°

(c) 35°

is the transversal.

Now, in $∆OCD$, we have:

#### Question 39:

In the given figure, AB || DC, BAD = 90°, ∠CBD = 28° and ∠BCE = 65°. Then ∠ABD = ?
(a) 32°
(b) 37°
(c) 43°
(d) 53°

Side DC of triangle BCD is produced to E.
$\therefore \angle BCE=\angle CBD+\angle BDC\phantom{\rule{0ex}{0ex}}⇒65°=28°+\angle BDC\phantom{\rule{0ex}{0ex}}⇒\angle BDC=37°$
Now,
is the transversal.

#### Question 40:

For what value of x, shall l || m?
(a) x = 50
(b) x = 70
(c) x = 60
(d) x = 45

(a) x = 50

for l || m we have,

#### Question 41:

For what value of x, shall l || m?
(a) x = 35
(b) x = 30
(c) x = 25
(d) x = 20

(c) x = 25

for l || m we have,

#### Question 42:

In the given figure, sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 35°
(b) 45°
(c) 55°
(d) 65°

(d) 65°

We have :

Side AB of triangle ABC is produced to E.
$\therefore \angle CAE=\angle ABC+\angle ACB\phantom{\rule{0ex}{0ex}}⇒135°=70°+\angle ACB\phantom{\rule{0ex}{0ex}}⇒\angle ACB=65°$

#### Question 43:

In ABC, BDAC, ∠CAE = 30° and ∠CBD = 40°. Then, ∠AEB = ?
(a) 70°
(b) 80°
(c) 50°
(d) 60°

(b) 80°

In $∆BCD$, we have:

Side CE of triangle AEC is produced to B.
$\therefore \angle AEB=\angle EAC+\angle ACE\phantom{\rule{0ex}{0ex}}⇒\angle AEB=30°+50°\phantom{\rule{0ex}{0ex}}⇒\angle AEB=80°$

#### Question 44:

In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, x = ?
(a) 108°
(b) 126°
(c) 162°
(d) 63°

(b) 126°
Let
Let the transversal intersect AB at P, CD at O and EF at Q.

Then, we have:

Also,
$\angle APO+\angle COP=180°\phantom{\rule{0ex}{0ex}}⇒x+54°=180°\phantom{\rule{0ex}{0ex}}⇒x=126°$

#### Question 45:

In the given figure, AB || CD || EF, EAAB and BDE is the transversal, such that DEF = 55°. Then, ∠AEB = ?
(a) 35°
(b) 45°
(c) 25°
(d) 55°

(a) 35°

We have :
$EA\perp AB$
Therefore, $\angle BAE=90°$
And AB  || EF and BE is transversal

$⇒\angle AEB+\angle BEF=90°\phantom{\rule{0ex}{0ex}}⇒\angle AEB+\angle DEF=90°\phantom{\rule{0ex}{0ex}}⇒\angle AEB+55°=90°\phantom{\rule{0ex}{0ex}}⇒\angle AEB=35°$

#### Question 46:

In the given figure, AMBC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?
(a) 20°
(b) 25°
(c) 15°
(d) 30°

(b) 25°

In $∆ABC$, we have:

In $∆ABM$, we have:

#### Question 47:

An exterior angle of a triangle is 110° and one of its interior opposite angles is 45°. The other interior opposite angle is
(a) 45°
(b) 65°
(c) 25°
(d) 135°

Figure

Let the required angle be $x°$.
We know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.
Then,
$110°=45°+x°\phantom{\rule{0ex}{0ex}}⇒x°=65°$
Therefore, the other interior opposite angle is 65o

#### Question 48:

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively, as shown in the figure, forming exterior angles ∠ACD, ∠BAE and ∠CBF. Then, ∠ACD + ∠BAE + ∠CBF = ?
(a) 240°
(b) 300°
(c) 320°
(d) 360°

(d) 360°

We have :

#### Question 49:

The angles of a triangle are in the ration 3 : 5 : 7. The triangle is
(a) acute-angled
(b) right-angled
(c) obtuse-angled
(d) isosceles

(a) acute-angled

Let the angles measure .
Then,
$3x+5x+7x=180°\phantom{\rule{0ex}{0ex}}⇒15x=180°\phantom{\rule{0ex}{0ex}}⇒x=12°$
Therefore, the angles are .
Hence, the triangle is acute-angled.

#### Question 50:

If the vertical angle of a triangle is 130°, then the angle between the bisectors of the base angles of the triangle is
(a) 65°
(b) 100°
(c) 130°
(d) 155°

(d) 155°

Let ABC be a triangle and let $\angle A=130°$.
Then, from $∆ABC$, we have:

Now, in $∆BOC$, we have:

#### Question 51:

The sides BC, BA and CA of ABC have been produced to D, E and F, respectively, as shown in the given figure. Then, ∠B = ?
(a) 35°
(b) 55°
(c) 65°
(d) 75°

(d) 75°

We have:

Side CD of triangle ABC is produced to D.
$\therefore \angle ACD=\angle BAC+\angle ABC\phantom{\rule{0ex}{0ex}}⇒110°=35°+\angle ABC\phantom{\rule{0ex}{0ex}}⇒\angle ABC=75°\phantom{\rule{0ex}{0ex}}⇒\angle B=75°$

#### Question 52:

In the adjoining figure, y = ?
(a) 36°
(b) 54°
(c) 63°
(d) 72°

(b) 54°

We have:

Also,

#### Question 53:

Assertion: If two angles of a triangle measure 50° and 70°, its third angle is 60°.
Reason: The sum of the angles of a triangle is 180°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:
Let the third angle be $x°$.
Then,

Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Since, both the Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 54:

Assertion: If a ray $\stackrel{\to }{CD}$ stands on a line $\stackrel{\to }{AB}$, such that ∠ACD = ∠BCD, then ∠ACD = 90°.
Reason: If a ray
$\stackrel{\to }{CD}$ stands on a line $\stackrel{\to }{AB}$, then ACD + ∠BCD = 180°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Assertion:

Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true and Reason is a correct explanation of Assertion.

#### Question 55:

Assertion: If the side BC of ABC is produced to D, then ∠ACD = ∠A + ∠B.
Reason: The sum of the angles of a triangle is 180°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Assertion:

Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true but Reason is not a correct explanation of Assertion.

#### Question 56:

Assertion: If two lines AB and CD intersect at O, such that AOC = 40°, then ∠BOC = 140°.

Reason: If two straight lines intersect each other, then vertically-opposite angles are equal.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.

Assertion:

Hence, the Assertion is true.
Reason:
Clearly, the given statement is true.
Hence, both the Assertion and Reason are true but Reason is not a correct explanation of Assertion.

#### Question 57:

Assertion: If AB || CD and t is the transversal as shown, then ∠3 = ∠5.

Reason: If a ray stands on a straight line, then the sum of the adjacent angles so formed is 180°.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(d) Assertion is false and Reason is true.

Assertion:

If the sum of two angles is 180o, it is not necessary that they are equal.
Hence, the Assertion is false.
Reason: "If a ray stands on a straight line, then the sum of the adjacent angles so formed is 180°."
Clearly, the given statement is true.

#### Question 58:

Match the following columns:

 Column I Column II (a) If x° and y° be the measures of two complementary angles, such that 2x = 3y, then x = ....... (p) 45° (b) If an angle is the supplement of itself, then the measure of the angle is ....... (q) 60° (c) If an angle is the complement of itself, then the measure of the angle is ....... (r) 54° (d) If x° and y° be the angles forming a linear pair, such that x − y = 60°, then = ........ (s) 90°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a)
We have :

Hence, (a)$\mathbf{⇒}$(r).

(b)
We have :
$x+x=180°\phantom{\rule{0ex}{0ex}}⇒2x=180°\phantom{\rule{0ex}{0ex}}⇒x=90°$
Hence, (b)$\mathbf{⇒}$(s).

(c)
We have :
$x+x=90°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, (c)$\mathbf{⇒}$(p).

(d)
We have:

Hence, (d)$\mathbf{⇒}$(q).

#### Question 59:

Match the following columns:

 Column I Column II (a) In the given figure, ACB is a straight line. Then, ∠ACD = ...... (p) 110° (b) In the given figure, ∠AOC = (2x − 10)° and ∠BOC = (3x − 10)°. Then, ∠AOD = ...... . (q) 85° (c) In the given figure, side BC of ∆ABC has been produced to D. If ∠A = 65° and ∠B = 60°, then ∠ACD = ? (r) 72° (d) In the given figure, if AB || DE, ∠CDE = 50° and ∠BAC = 45°, then ∠ACB = ...... . (s) 125°
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a)
We have:

Hence, (a)$\mathbf{⇒}$(r).

(b)
We have:
$2x-10+3x-10=180°\phantom{\rule{0ex}{0ex}}⇒5x=200°\phantom{\rule{0ex}{0ex}}⇒x=40°\phantom{\rule{0ex}{0ex}}\therefore \angle BOC=\left[3×40°-10\right]=110°$
Also,

Hence, (b)$\mathbf{⇒}$(p)

(c)
We have :

Hence, (c)$\mathbf{⇒}$(s).

(d)
is the transversal.

From $∆ABC$, we have:

Hence, (d)$\mathbf{⇒}$(q)

#### Question 1:

The angles of a triangle are in the ratio 3 : 2 : 7. Find the measure of each of its angles.

Let the angles measure .
Then,

Therefore, the measures of its angles are .

#### Question 2:

In ABC, if ∠A − ∠B = 40° and ∠B − ∠C = 10°, find the measures of ∠A, ∠B and ∠C.

Now,

#### Question 3:

The side BC of ABC has been increased on both sides as shown. If ∠ABD = 105° and ∠ACE = 110°, find ∠A.

We have:

Side BC of triangle ABC is produced to D.

#### Question 4:

Prove that the bisectors of two adjacent supplementary angles include a right angle.

Let AOB denote a straight line and let  be the supplementary angles.
Then, .
Let .
Then, we have:
$\angle AOE=\angle COE=\frac{1}{2}x°\phantom{\rule{0ex}{0ex}}\angle BOF=\angle COF=\frac{1}{2}\left(180-x\right)°$
Now,
$\angle EOF=\angle COE+\angle COF$
$=\frac{1}{2}x°+\frac{1}{2}\left(180-x°\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(x+180-x\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×180°\phantom{\rule{0ex}{0ex}}=90°$
Hence, the bisectors of two supplementary angles include a right angle.

#### Question 5:

If one angle of a triangle is equal to the sum of the two other angles, show that the triangle is right-angled.

Let ABC be a triangle and let $\angle A=\angle B+\angle C$.
Then,

Therefore, the triangle is right-angled at A.

#### Question 6:

In the given figure, ACB is a straight line and CD is a line segment, such that ACD = (3x − 5)° and ∠BCD = (2x + 10)°. Then, x = ?
(a) 25
(b) 30
(c) 35
(d) 40

(c) 35
We have:

#### Question 7:

In the given figure, AOB is a straight line. If AOC = 40°, ∠COD = 4x° and ∠BOD = 3x°, then x = ?
(a) 20
(b) 25
(c) 30
(d) 35

(a) 20

We have:

#### Question 8:

The supplement of an angle is six times its complement. The measure of this angle is
(a) 36°
(b) 54°
(c) 60°
(d) 72°

(d) 72°

Let the required angle be $x°$.
Then, measure of its supplement = $\left(180-x\right)°$ and
measure of its complement = $\left(90-x\right)°$.
$\therefore 180-x=6\left(90-x\right)\phantom{\rule{0ex}{0ex}}⇒5x=360°\phantom{\rule{0ex}{0ex}}⇒x=72°$

#### Question 9:

In the given figure, AB || CD || EF. If ABC = 85°, ∠BCE = x° and CEF = 130°, then x = ?
(a) 30
(b) 25
(c) 35
(d) 15

(c) 35

is the transversal.

Also,  is the transversal.

#### Question 10:

In the given figure, AB || CD, BAD = 30° and ∠ECD = 50°. Find ∠CED.

is the transversal.

In $∆CDE$, we have:

#### Question 11:

In the given figure, BAD || EF, AEF = 55° and ∠ACB = 25°. Find ∠ABC.

is the transversal.

Also,

From $∆ABC$, we have:

#### Question 12:

In the given figure, BE AC, ∠DAC = 30° and ∠DBE = 40°. Find ∠ACB and ∠ADB.

From $∆BEC$, we have:

Side CD of triangle ACD is produced to B.

#### Question 13:

In the given figure, AB || CD and EF is a transversal, cutting them at G and H, respectively. If EGB = 35° and QPEF, find the measure of ∠PQH.

We have:

Also,

Now, in $∆PQH$, we have:

#### Question 14:

In the given figure, AB || CD and EF AB. If EG is the transversal, such that ∠GED = 130°, find ∠EGF.

We have:

is the transversal.

#### Question 15:

Match the following columns:

 Column I Column II (a) An angle is 10° more than its complement. The measure of the angle is ...... . (p) 160° (b) If in ∆ABC, ∠A = 65° and ∠B − ∠C = 25°, then ∠B = ...... (q) 50° (c) In ∆ABC, ∠A = 40° and ∠B = ∠C. If EF || BC, then ∠EFC = ...... . (r) 70° (d) If the angles around a point are 2x°, 3x°, 5x° and 40°, the measure of the largest angle is ...... (s) 110°
(a) ......
(b) ......
(c) ......
(d) ......

(a)
Let the required angle be $x°$.
Then, measure of its complement = $\left(90-x\right)°$.
$\therefore x=\left(90-x\right)+10\phantom{\rule{0ex}{0ex}}⇒2x=100°\phantom{\rule{0ex}{0ex}}⇒x=50°$
Hence, (a)$⇒$(q).

(b)
We have:

Hence, (b)$⇒$(r).

(c)
In $∆ABC$, we have:

is the transversal.

Now, side AF of triangle AEF is produced to C.
$\therefore \angle EFC=\angle EAF+\angle AEF\phantom{\rule{0ex}{0ex}}⇒\angle EFC=40°+70°\phantom{\rule{0ex}{0ex}}⇒\angle EFC=110°$
Hence, (c)$⇒$(s).

(d)
We know that the sum of the angles around a point is $360°$.
$\therefore 2x+3x+5x+40°=360°\phantom{\rule{0ex}{0ex}}⇒10x=320°\phantom{\rule{0ex}{0ex}}⇒x=32°$
Hence, the angles are

Therefore, the largest angle is $160°$.
Hence, (d)$⇒$(p).

#### Question 16:

(i) In the given figure, lines AB and CD intersect at O, such that AOD + ∠BOD + ∠BOC = 300°. Find ∠AOD.

(ii) In the given figure, AB || CD, ∠APQ = 50° and ∠PRD = 120°. Find ∠QPR.

(i)
We know that the sum of the angles around a point is $360°$.
$\therefore \angle AOD+\angle BOD+\angle BOC+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒\left(\angle AOD+\angle BOD+\angle BOC\right)+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒300°+\angle AOC=360°\phantom{\rule{0ex}{0ex}}⇒\angle AOC=60°$
Also,

Now,

(ii)
is the transversal.

Side QR of triangle PQR is produced to D.
$\therefore \angle PRD=\angle QPR+\angle PQR\phantom{\rule{0ex}{0ex}}⇒120°=\angle QPR+50°\phantom{\rule{0ex}{0ex}}⇒\angle QPR=70°$

#### Question 17:

In the given figure, BE is the bisector of B and CE is the bisector of ∠ACD. Prove that $\angle BEC=\frac{1}{2}\angle A.$

Side BC of triangle ABC is produced to D.

Also, side BC of triangle EBC is produced to D.

From (i) and (ii), we get:
$\frac{1}{2}\angle B+\angle BEC=\frac{1}{2}\angle B+\frac{1}{2}\angle A\phantom{\rule{0ex}{0ex}}⇒\angle BEC=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\angle }\mathbit{A}$

#### Question 18:

In ABC, sides AB and AC are produced to D and E, respectively. BO and CO are the bisectors of ∠CBD and ∠BCE, respectively. Prove that $\angle BOC=90°-\frac{1}{2}\angle A.$

Let

We have:

Also,

Now, in $∆OBC$, we have:

#### Question 19:

Of the three angles of a triangle, one is twice the smallest and another is thrice the smallest. Find the angles.

Let ABC be a triangle and let $\angle C$ be the smallest angle.
Then, let
Now,

#### Question 20:

In ABC, ∠B = 90° and BDAC. Prove that ∠ABD = ∠ACB.

From the right $∆ABC$, we have: