Rs Aggarwal 2018 Solutions for Class 9 Math Chapter 8 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 9 students for Math Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 9 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### 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:

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 13:

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 14:

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 15:

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 16:

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 17:

In the figure given alongside, AB || CD, EF || BC, ∠BAC = 60º and ∠DHF = 50º. Find ∠GCH and ∠AGH. 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

#### 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 the given figure, AMBC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ? In $∆ABC$, we have:

In $∆ABM$, we have:

#### Question 23:

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

#### 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, AB || DE and BD || FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y. 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.

#### Question 26:

In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, find the value of x. 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°.

#### Question 27:

In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO. 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°.

#### Question 28:

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

#### Question 29:

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

#### Question 1:

In ∆ABC, 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

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).

#### Question 2:

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

Figure

(c) 53°

Let

#### Question 3:

In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30° $\therefore \angle A+\angle B=\angle ACD\phantom{\rule{0ex}{0ex}}⇒\angle A+50°=110°\phantom{\rule{0ex}{0ex}}⇒\angle A=60°$
Hence, the correct answer is option (b).

#### Question 4:

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

We have :

Also,

#### Question 5:

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

We have :

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

#### Question 6:

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

We have :

#### Question 7:

The the given figure, EAD  BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ∆BAC, BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is
(a) 20
(b) 25
(c) 30
(d) 35 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).

#### Question 8:

In the given figure, two rays BD and CE intersect at a point A. The side BC of ∆ABC have been produced on both sides to points F and G respectively. If ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?
(a) x + y – 180
(b) x + y + 180
(c) 180 – (x + y)
(d) x + y + 360° 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).

#### Question 9:

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ∠OAE = x° and ∠DBF = y°. If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?
(a) 190°
(b) 230°
(c) 210°
(d) 270°

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).

#### Question 10:

In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E then ∠ECD = ?
(a) 60°
(b) 50°
(c) 40°
(d) 25° 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).

#### Question 11:

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 12:

In the given figure, side BC of ∆ABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠CBD = 7y°. Then, the value of x is (a) 60
(b) 50
(c) 45
(d) 35

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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