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#### Page No 252:

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

Hence, the measures of the angles are .

#### Page No 252:

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

Therefore,

#### Page No 252:

Let .

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

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

#### Page No 252:

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

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

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

#### Page No 252:

Then,

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

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

#### Page No 252:

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:

#### Page No 252:

Let .
Then,

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

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

#### Page No 252:

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,

#### Page No 252:

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

Hence, .

#### Page No 252:

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.

#### Page No 252:

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.

#### Page No 252:

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.

#### Page No 252:

Side BC of triangle ABC is produced to D.

Also, in triangle ABC,

#### Page No 252:

Side BC of triangle ABC is produced to D.

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

And,

#### Page No 253:

(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

#### Page No 253:

In the given figure, AB || CD and AC is the transversal.

∴ ∠ACD = ∠BAC = 60º     (Pair of alternate angles)

Or ∠GCH = 60º

Now, ∠GHC = ∠DHF = 50º      (Vertically opposite angles)

In ∆GCH,

∠AGH = ∠GCH + ∠GHC       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ ∠AGH = 60º + 50º = 110º

#### Page No 253:

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)

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

#### Page No 254:

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:

#### Page No 254:

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.

#### Page No 254:

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

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

#### Page No 254:

In $∆ABC$, we have:

In $∆ABM$, we have:

#### Page No 254:

In the given figure, EF || BD and CE is the transversal.

∴ ∠CAD = ∠AEF         (Pair of corresponding angles)

In ∆ABC,

∠CAD = ∠ABC + ∠ACB       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ 55° = ∠ABC + 25°

⇒ ∠ABC = 55° − 25° = 30°

Thus, the measure of ∠ABC is 30°.

#### Page No 255:

Let
From $∆ABC$, we have:

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

#### Page No 255:

We have, ∠FGE + ∠FGH = 180°       (Linear pair of angles)

y + 120° = 180°

⇒ y = 180° − 120° = 60°

Now, AB || DF and BD is the transversal.

∴ ∠ABD = ∠BDF           (Pair of alternate angles)

⇒ ∠BDF = 50°

Also, BD || FG and DF is the transversal.

∴ ∠BDF = ∠DFG           (Pair of alternate angles)

⇒ ∠DFG = 50°       .....(1)

In ∆EFG,

∠FGH = ∠EFG + ∠FEG       (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)

⇒ 120° = 50° + x                   [Using (1)]

⇒ x = 120° − 50° = 70°

Thus, the values of x and y are 70° and 60°, respectively.

#### Page No 255:

It is given that, AB || CD and EF is a transversal.

∴ ∠EFD = ∠AEF          (Pair of alternate angles)

⇒ ∠EFD = 65°

⇒ ∠EFG + ∠GFD = 65°

⇒ ∠EFG + 30° = 65°

⇒ ∠EFG = 65° − 30° = 35°

In ∆EFG,

∠EFG + ∠GEF + ∠EGF = 180°           (Angle sum property)

⇒ 35° + x + 90° = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the value of x is 55°.

#### Page No 255:

In the given figure, AB || CD and AE is the transversal.

∴ ∠DOE = ∠BAE         (Pair of corresponding angles)

⇒ ∠DOE = 65°

In ∆COE,

∠DOE = ∠OEC + ∠ECO          (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ 65° = 20° + ∠ECO

⇒ ∠ECO = 65° − 20° = 45°

Thus, the measure of ∠ECO is 45°.

#### Page No 255:

In the given figure, AB || CD and EF is a transversal.

∴ ∠PHQ = ∠EGB          (Pair of alternate exterior angles)

⇒ ∠PHQ = 35°

In ∆PHQ,

∠PHQ + ∠QPH + ∠PQH = 180°     (Angle sum property)

⇒ 35° + 90° + x = 180°

⇒ 125° + x = 180°

x = 180° − 125° = 55°

Thus, the measure of ∠PQH is 55°.

#### Page No 255:

In the given figure, AB || CD and GE is the transversal.

∴ ∠GED + ∠EGF = 180°       (Sum of adjacent interior angles on the same side of the transversal is supplementary)

⇒ 130° + ∠EGF = 180°

⇒ ∠EGF = 180° − 130° = 50°

Thus, the measure of ∠EGF is 50°.

#### Page No 257:

LCM of 3, 4 and 6 = 12

3∠A = 4∠B = 6∠C      (Given)

Dividing throughout by 12, we get

$\frac{3\angle \mathrm{A}}{12}=\frac{4\angle \mathrm{B}}{12}=\frac{6\angle \mathrm{C}}{12}$

$⇒\frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}$

Let $\frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}=k$, where k is some constant

Then, ∠A = 4k,  ∠B = 3k, ∠C = 2k

∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2

Hence, the correct answer is option (b).

(c) 53°

Let

#### Page No 258:

$\therefore \angle A+\angle B=\angle ACD\phantom{\rule{0ex}{0ex}}⇒\angle A+50°=110°\phantom{\rule{0ex}{0ex}}⇒\angle A=60°$
Hence, the correct answer is option (b).

(d) 75°

We have :

Also,

#### Page No 258:

(a) 65°

We have :

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

(d) 360°

We have :

#### Page No 258:

In the given figure, ∠CAD = ∠EAF            (Vertically opposite angles)

In ∆ABD,

⇒ (x + 10)° + (x° + 30°) + 90° = 180°

⇒ 2x° + 130° = 180°

⇒ 2x° = 180° − 130° = 50°

⇒ x = 25

Thus, the value of x is 25.

Hence, the correct answer is option (b).

#### Page No 259:

In the given figure, ∠ABF + ∠ABC = 180°         (Linear pair of angles)

∴ x° + ∠ABC = 180°

⇒ ∠ABC = 180° − x°        .....(1)

Also, ∠ACG + ∠ACB = 180°         (Linear pair of angles)

∴ y° + ∠ACB = 180°

⇒ ∠ACB = 180° − y°        .....(2)

Also, ∠BAC = ∠DAE = z°       .....(3)         (Vertically opposite angles)

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180°       (Angle sum property)

∴ z° + 180° − x° + 180° − y° = 180°        [Using (1), (2) and (3)]

⇒ zxy − 180

Hence, the correct answer is option (a).

#### Page No 259:

In the given figure, ∠BOD = ∠COA          (Vertically opposite angles)

∴ ∠BOD = 40°        .....(1)

In ∆ACO,

∠OAE = ∠OCA + ∠COA        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ x° = 80° + 40° = 120°           .....(2)

In ∆BDO,

∠DBF = ∠BDO + ∠BOD        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ y° = 70° + 40° = 110°           [Using (1)]         .....(3)

Adding (2) and (3), we get

x° + y° = 120° + 110° = 230°

Hence, the correct answer is option (b).

#### Page No 259:

Let
Then,

Hence, the angles are

Side BC of triangle ABC is produced to E.
$\therefore \angle ACE=\angle A+\angle B\phantom{\rule{0ex}{0ex}}⇒\angle ACD+\angle ECD=90°+60°\phantom{\rule{0ex}{0ex}}⇒90°+\angle ECD=150°\phantom{\rule{0ex}{0ex}}⇒\angle ECD=60°$
Hence, the correct answer is option (a).

#### Page No 259:

(c) 115°

In $∆ABC$, we have:

In $∆OBC$, we have:

#### Page No 260:

Disclaimer: In the question ACD should be 7y°.

In the given figure, ∠ACB + ∠ACD = 180°         (Linear pair of angles)

∴ 5y° + 7y° = 180°

⇒ 12y° = 180°

⇒ y = 15         .....(1)

In ∆ABC,

∠A + ∠B + ∠ACB = 180°       (Angle sum property)

∴ 3y° + x° + 5y° = 180°

⇒ x° + 8y° = 180°

⇒ x° + 8 × 15° = 180°              [Using (1)]

⇒ x° + 120° = 180°

⇒ x° = 180° − 120° = 60°

Thus, the value of x is 60.

Hence, the correct answer is option (a).

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