RS AGGARWAL CLASS 9 CHAPTER 8 TRIANGLES EXERCISE 8

 EXERCISE 8

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Q1 | Ex-8 |Class 9 | RS AGGARWAL | TRIANGLES |Ch-8 |myhelper

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Question 1:

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

Answer 1:

In ∆ABC,∠A+∠B+∠C=180° [Sum of the angles of a triangle]

⇒∠A+76°+48°=180°

⇒∠A+124°=180°

⇒∠A=56°

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Question 2:

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

Answer 2:

Let the angles of the given triangle measure (2x)°, (3x)° and (4x)°, respectively.
Then,
2x+3x+4x=180° [Sum of the angles of a triangle]
⇒9x=180°
⇒x=20°
Hence, the measures of the angles are 2×20°=40°, 3×20°=60° and 4×20°=80°.

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Question 3:

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

Answer 3:

Let 3∠A=4∠B=6∠C=x°.
Then,

$\angle A=\left(\frac{x}{3}\right)^{\circ}$ ,$\angle B=\left(\frac{x}{4}\right)^{\circ}$  , $\angle C=\left(\frac{x}{6}\right)^{\circ}$ 

∴ $\frac{x}{3}+\frac{x}{4}+\frac{x}{6}=180^{\circ}$ [Sum of the angles of a triangle]

⇒4x+3x+2x=2160°
⇒9x=2160°
⇒x=240°

Therefore,

$\angle A=\left(\frac{240}{3}\right)^{\circ}$=80°

$\angle B=\left(\frac{240}{4}\right)^{\circ}$=60°

$\angle C=\left(\frac{240}{6}\right)^{\circ}$=40°

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Question 4:

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

Answer 4:

Let ∠A+∠B=108° and ∠B+∠C=130°.
⇒∠A+∠B+∠B+∠C=(108+130)°

⇒(∠A+∠B+∠C)+∠B=238° [∵∠A+∠B+∠C=180°]

⇒180°+∠B=238°

⇒∠B=58°

∴ ∠C=130°−∠B
=(130−58)°=72°

∴ ∠A=108°−∠B
=(108−58)°=50°

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Question 5:

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

Answer 5

Let ∠A+∠B=125° and ∠A+∠C=113°.
Then,
∠A+∠B+∠A+∠C=(125+113)°

⇒(∠A+∠B+∠C)+∠A=238°

⇒180°+∠A=238°

⇒∠A=58°

∴ ∠B=125°−∠A
=(125−58)°=67°

∴∠C=113°− ∠A
=(113−58)°=55°

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Question 6:

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

Answer 6

Given: ∠P−∠Q=42° and ∠Q−∠R=21°
Then,
∠P=42°+∠Q and ∠R=∠Q−21°

∴42°+∠Q+∠Q+∠Q−21°=180° [Sum of the angles of a triangle]

⇒3∠Q=159°

⇒∠Q=53°

∴∠P=42°+∠Q
=(42+53)°=95°

∴∠R=∠Q−21°
=(53−21)°=32°

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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.

Answer 7:

Let ∠A+∠B=116° and ∠A−∠B=24°
Then,
∴∠A+∠B+∠A−∠B=(116+24)°

⇒2∠A=140°

⇒∠A=70°

∴∠B=116°−∠A
=(116−70)°=46°

Also, in ∆ ABC:
∠A+∠B+∠C=180° [Sum of the angles of a triangle]

⇒70°+46°+∠C=180°

⇒∠C=64°

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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.

Answer 8:

Let ∠A=∠B and ∠C=∠A+18°.
Then,
∠A+∠B+∠C=180° [Sum of the angles of a triangle]∠A+∠A+∠A+18°=180°

⇒3∠A=162°

⇒∠A=54°
Since,
∠A=∠B⇒∠B=54°

∴∠C=∠A+18°

=(54+18)°=72°

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Question 9:

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

Answer 9:

Let the smallest angle of the triangle be ∠C and let ∠A=2∠C and ∠B=3∠C.
Then,
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 2\angle C+3\angle C+\angle C=180° \\ \Rightarrow 6\angle =180° \\ \Rightarrow \angle C=30° \end{gathered}$$

$\therefore \angle A=2\angle C$
         $$\begin{gathered} =2\left(30\right)° \\ =60° \end{gathered}$$
Also,
$$\begin{gathered} \angle B=3\angle C \\ =3\left(30\right)° \\ =90° \end{gathered}$$

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Question 10:

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

Answer 10:

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

Hence, .

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Question 11:

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

Answer 11:

Let ABC be a triangle.
Then,

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

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Question 12:

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

Answer 12:

Let ABC be the triangle.
Let 
Then,


Also, let 
Then,


And let
Then,


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

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Question 13:

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

Answer 13:

Let ABC be a triangle and let .
Then, we have:

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

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Question 14:

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

Answer 14:

Side BC of triangle ABC is produced to D.

Also, in triangle ABC,

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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.

Answer 15:

Side BC of triangle ABC is produced to D.



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








And,

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Question 16:

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

Answer 16:

(i)
Side AC of triangle ABC is produced to E.
∴∠EAB=∠B+∠C

⇒110°=x+∠C ...(i)

Also,
∠ACD+∠ACB=180° [linear pair]

⇒120°+∠ACB=180°

⇒∠ACB=60°

⇒∠C=60°
Substituting the value of ∠C in (i), we get x=50

(ii)
From ∆ABC we have:
∠A+∠B+∠C=180° [Sum of the angles of a triangle]

⇒30°+40°+∠C=180°

⇒∠C=110°

⇒∠ACB=110°

Also,
∠ECB+∠ECD=180°

⇒110+∠ECD=180°

⇒∠ECD=70°

Now in ΔECD,

∴∠AED=∠ECD+∠EDC [exterior angle property]

⇒x=70°+50°

⇒x=120°

(iii)

∠ACB+∠ACD=180° [Linear pair]

⇒∠ACB+115°=180°

⇒∠ACB=65°

Also, 

∠EAF=∠BAC [Vertically Opposite Angle]

⇒BAC=60°

∴∠BAC+∠ABC+∠ACB=180° [Sum of the angles of a triangle]

⇒60°+x+65°=180°

⇒x=55°

(iv)

∠BAE=∠CDE [Alternate angles]

⇒∠CDE=60°

∴∠ECD+∠CDE+∠CED=180° [Sum of the angles of a triangle]

⇒45°+60°+x=180°

⇒x=75°

(v)

From ΔABC, we have

∠BAC+∠ABC+∠ACB=180° [Sum of the angles of a triangle]

⇒40°+∠ABC+90°=180°

⇒∠ABC=50°

Also, from EBD, we have

⇒∠BED+∠EBD+∠BDE=180° [Sum of the angles of a triangle]

⇒100°+50°+x=180°

⇒x=30°

(vi)

From ΔABE , we have

∠BAE+∠ABE+∠AEB=180° [Sum of the angles of a triangle]

⇒75°+65°+AEB=180°

⇒∠AEB=40°

∴∠AEB=∠CED [Vertically Opposite angles]

⇒∠CED=40°

Also, Form ΔCDE, we have

∠ECD+∠CDE+∠CED=180°  [Sum of the angles of a triangle]

⇒110°+x+40°=180°

⇒x=30°

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Question 17:

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

Answer 17:

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
 

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Question 18:

Calculate the value of x in the given figure.

Answer 18:

Join A and D to produce AD to E.
Then,
$$\begin{gathered} \angle CAD+\angle DAB=55° \text{and} \\ \angle CDE+\angle EDB=x° \end{gathered}$$
Side AD of triangle ACD is produced to E.
$\therefore \angle CDE=\angle CAD+\angle ACD ...\left(i\right)$   (Exterior angle property)
Side AD of triangle ABD is produced to E.
$\therefore \angle EDB=\angle DAB+\angle ABD ...\left(ii\right)$   (Exterior angle property)
$$\begin{gathered} \mathrm{Adding} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)\mathrm{we} \mathrm{get}, \\ \angle CDE+\angle EDB=\angle CAD+\angle ACD+\angle DAB+\angle ABD \end{gathered}$$
 
                 $$\begin{gathered} \Rightarrow x°=\left(\angle CAD+\angle DAB\right)+30°+45° \\ \Rightarrow x°=55°+30°+45° \\ \Rightarrow x°=130° \\ \Rightarrow x=130 \end{gathered}$$

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Question 19:

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

Answer 19:

∠BAC+∠CAE=180° [∵BE is a straight line]

⇒∠BAC+108°=180°

⇒∠BAC=72°

∵BE is a straight line

⇒∠BAC+108°=180°

⇒∠BAC=72°

Now, divide 72°72° in the ratio 1 : 3.
∴a+3a=72°

⇒a=18°

∴a=18° and 3a=54°

Hence, the angles are 18° and 54°

∴∠BAD=18° and ∠DAC=54°

∴∠BAD=18° and ∠DAC=54°



Given,

AD=DB⇒∠DAB=∠DBA=18°

In ∆ABC, we have:
∠BAC+∠ABC+∠ACB=180° [Sum of the angles of a triangle]

⇒72°+18°+x°=180°

⇒x°=90°

∴ x=90

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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.

Answer 20:

Side BC of triangle ABC is produced to D.
∠ACD=∠B+∠A ...(i)
Side AC of triangle ABC is produced to E.
∠BAC=∠B+∠C ...(i)
And side AB of triangle ABC is produced to F.
∠CBF=∠C+∠A ...(iii)

Adding (i), (ii) and (iii), we get:∠ACD+∠BAE+∠CBF=2(∠A+∠B+∠C)

Adding (i), (ii) and (iii), we get:∠ACD+∠BAE+∠CBF=2∠A+∠B+∠C
=2(180)°

=360°=4×90°

=4 right angles

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

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Question 21:

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

Answer 21:

In ∆ACE, we have :
∠A+∠C+∠E=180° ...(i) [Sum of the angles of a triangle]
In ∆BDF∆BDF, we have :
∠B+∠D+∠F=180° ...(ii) [Sum of the angles of a triangle]​

Adding (i) and (ii), 

we get:∠A+∠C+∠E+∠B+∠D+∠F=(180+180)°

⇒∠A+∠B+∠C+∠D+∠E+∠F=360°

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Question 22:

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?

Answer 22:

In ∆ABC, we have:
∠A+∠B+∠C=180° [Sum of the angles of a triangle]

⇒∠A+70°+20°=180°

⇒∠A=90°

⇒$\frac{1}{2}$∠A=45°

⇒∠BAN=45°

In ∆ABM, we have:
∠ABM+∠AMB+∠BAM=180° [Sum of the angles of a triangle]

⇒70°+90°+∠BAM=180°

⇒∠BAM=20°

∴∠MAN=∠BAN−∠BAM

⇒∠MAN=45°−20°

⇒∠MAN=25°

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Question 23:

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

Answer 23:

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

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

⇒ ∠CAD = 55°

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°.

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Question 24:

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

Answer 24:

Let ∠A=(3x)°, ∠B=(2x)° and ∠C=x°
From ∆ABC, we have:
∠A+∠B+∠C=180° [Sum of the angles of a triangle]

⇒3x+2x+x=180°

⇒6x=180°

⇒x=30°

∴∠A=330°=60° 

 ∠B=230°=60° and ∠C=30°

Side BC of triangle ABC is produced to E.
∴∠ACE=∠A+∠B

⇒∠ACD+∠ECD=(90+60)°

⇒90+∠ECD=150°

⇒∠ECD=60°

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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.

Answer 25:

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.

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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.

Answer 26:

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°.

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Question 27:

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

Answer 27:

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°.

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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.

Answer 28:

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°.

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Question 29:

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

Answer 29:

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°.