Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-9.2, Triangle And Its Angles, Class 9, Maths

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1) The exterior angles, obtained on producing the base of a triangle both ways are 1040  and 1360. Find all the angles of the triangle.

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∠ACD = ∠ABC + ∠BAC [Exterior angle property]
Now∠ABC = 1800 − 1360 = 440 [Linera pair]
∠ACB = 1800 − 1040 = 760 [Linera pair]
Now,InΔABC
∠A + ∠ABC + ∠ACB = 180[Sum of all angles of a triangle]
⇒ ∠A + 44+ 76= 1800
⇒ ∠A = 1800 − 44− 760
⇒ ∠A = 600

 

Q2) In a triangle ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 1800.

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Let ∠ABD = 2x and ∠ACE = 2y
∠ABC = 180− 2x [Linera pair]
∠ACB = 180− 2y [Linera pair]
∠A + ∠ABC + ∠ACB = 180[Sum of all angles of a triangle]
⇒ ∠A + 1800 − 2x + 180− 2y = 1800
⇒ − ∠A + 2x + 2y = 1800
⇒ x + y = 900 + Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A

Now in ΔBQC x + y + ∠BQC = 1800 [Sum of all angles of a triangle]
⇒ 90+ Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A + ∠BQC = 1800
⇒ ∠BQC = 90Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A….(i)
and we know that∠BPC = 900 + Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A….(ii)
Adding (i) and (ii) we get ∠BPC + ∠BQC = 1800

Hence proved.

 

Q3) In figure 9.30, the sides BC, CA and AB of a triangle ABC have been produced to D, E and F respectively. If ∠ACD = 1050 and ∠EAF = 450, find all the angles of the triangle ABC.

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∠BAC = ∠EAF = 450 [Verticallyopposite angles]
∠ABC = 1050 − 450 = 600 [Exterior angle property]
∠ACD = 1800 − 1050 = 750 [Linear pair]

 

Q4) Compute the value of x in each of the following figures:

(i)

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

∠BAC = 1800 − 1200 = 600 [Linear pair]
∠ACB = 180− 1120 = 680 [Linear pair]
∴ x = 180− ∠BAC − ∠ACB
= 1800 − 600 − 680 = 520 [Sum of all angles of a triangle]

 

(ii)

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

∠ABC = 1800 − 120= 600 [Linear pair]
∠ACB = 180− 1100 = 700 [Linear pair]
∴ e∠BAC = x = 1800 − ∠ABC − ∠ACB
= 1800 − 60− 700 = 500 [Sum of all angles of a triangle]

 

(iii)

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

∠BAE = ∠EDC = 520 [Alternate angles]
∴∠DEC = x = 1800 − 400 − ∠EDC = 1800 − 400 − 52
= 1800 − 920 = 880 [Sum of all angles of a triangle]

 

(iv) 

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

CD is produced to meet AB at E.

∠BEC = 1800 − 45− 500 = 85[Sumofall angles of a triangle]
∠AEC = 180− 85= 950 [Linear pair]
∴x = 950 + 350 = 1300 [Exterior angle property] .

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q5) In figure 9.35, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Let∠BAD = Z,∠BAC = 3Z
⇒ ∠BDA = ∠BAD = Z           (∵AB = DB)
Now∠BAD + ∠BAC + 1080 = 1800 [Linear pair]
⇒ Z + 3Z + 108= 1800
⇒ 4Z = 720
⇒ Z = 180
Now, In ΔADC∠ADC + ∠ACD = 1080 [Exterior angle property]
⇒ x + 180 = 1800
⇒ x = 900

 

Q6) ABC is a triangle. The bisector of the exterior angle at B and the bisector of ∠C intersect each other at D. Prove that ∠D = Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A.

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Let ∠ABE = 2x and ∠ACB = 2y ∠ABC = 1800 − 2x [Linear pair]
∴∠A = 1800 − ∠ABC − ∠ACB [ angle sum property]
= 180− 1800 + 2x + 2y = 2(x − y)…..(i)
Now,∠D = 1800 − ∠DBC − ∠DCB
⇒ ∠D = 1800 − (x + 1800 − 2x) − y
⇒ ∠D = 1800 − x − 1800 + 2x − y = (x − y) = Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A…..from(i)
Hence,∠D = Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠A.

 

Q7) In figure 9.36, AC⊥CE and ∠A:∠B:∠C = 3:2:1,find

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

∠A:∠B:∠C = 3:2:1
Let the angles be 3x,2x and x
⇒ 3x + 2x + x = 1800 [ angle sum property]
⇒ 6x = 1800
⇒ x = 30= ∠ACB
∴∠ECD = 180− ∠ACB − 90[Linear pair]
= 1800 − 300 − 900 = 600
∴∠ECD = 600

  

Q8) In figure 9.37, AM⊥BC and AN is the bisector of ∠A. If ∠B = 650 and ∠C = 330,find ∠ MAN..

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Let∠BAN = ∠NAC = x [∵AN bisects ∠A]
∴∠ANM = x + 330 [Exterior angle property]
In ΔAMB ∠BAM = 90− 65= 250 [Exterior angle property]
∴∠MAN = ∠BAN − ∠BAM = (x − 25)0
Now in ΔMAN,(x − 25)+ (x + 33)+ 900 = 180[ angle sum property]
⇒ 2x + 80 = 900
⇒ 2x = 820
⇒ x = 410
∴MAN = x − 250 = 410 − 250 = 160

 

Q9) In a triangle ABC, AD bisects ∠A and ∠C> ∠B.Provethat ∠ADB > ∠ADC..

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∵∠C>∠B                               [Given]

⇒ ∠C + x>∠B + x                 [Adding x on both sides]

⇒  180° – ∠ ADC>180^{0} – ∠ ADB

⇒  −∠ ADC> – ∠ ADB

⇒  ∠ ADB > ∠ ADC

Hence proved.

 

Q10) In triangle ABC, BD⊥AC and CE⊥AB. If BD and CE intersect at O, prove that ∠BOC = 1800 − ∠A..

Solution:

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

In quadrilateral AEOD

∠A + ∠AEO + ∠EOD + ∠ADO = 3600
⇒ ∠A + 900 + 90+ ∠EOD = 3600
⇒ ∠A + ∠BOC = 1800 [∵∠EOD = ∠BOC  vertically opposite angles]
⇒ ∠BOC = 1800 − ∠A

 

Q11) In figure 9.38, AE bisects ∠CAD and ∠B = ∠C. Prove that AE∥BC.

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Let∠B = ∠C = x
Then,∠CAD = ∠B + ∠C = 2x (exterior angle)
Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics∠CAD = x
⇒ ∠EAC = x
⇒ ∠EAC = ∠C

These are alternate interior angles for the lines AE and BC

∴AE∥BC

 

Q12) In figure 9.39, AB∥DE.Find∠ACD.

Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Solution:

Since AB∥DE

∴∠ABC = ∠CDE = 400 [Alternate angles]
∴∠ACB = 1800 − ∠ABC − ∠BAC
= 1800 − 400 − 30= 1100
∴∠ACD = 1800 − 1100 [Linear pair] = 700

 

Q13) . Which of the following statements are true (T) and which are false (F) :

(i) Sum of the three angles of a triangle is 180°.

Solution: (i) T

(ii) A triangle can have two right angles.

Solution: (ii) F

(iii) All the angles of a triangle can be less than 60°.

Solution: (iii) F

(iv) All the angles of a triangle can be greater than 60°.

Solution: (iv) F

(v) All the angles of a triangle can be equal to 60°.

Solution: (v) T

(vi) A triangle can have two obtuse angles.

Solution: (vi) F

(vii) A triangle can have at most one obtuse angles.

Solution: (vii) T

(viii) If one angle of a triangle is obtuse, then it cannot be a right angled triangle.

Solution: (viii) T

(ix) An exterior angle of a triangle is less than either of its interior opposite angles.

Solution: (ix) F

(x) An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

Solution: (x) T

(xi) An exterior angle of a triangle is greater than the opposite interior angles.

Solution: (xi) T

 

Q14) Fill in the blanks to make the following statements true:

(i) Sum of the angles of a triangle is _______ . 

(ii) An exterior angle of a triangle is equal to the two ________ opposite angles.

(iii) An exterior angle of a triangle is always ________ than either of the interior opposite angles.

(iv) A triangle cannot have more than _______ right angles.

(v) A triangles cannot have more than _______ obtuse angles.

Solution:

(i) 1800

(ii) Interior

(iii) Greater

(iv) One

(v) One

The document Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
All you need of Class 9 at this link: Class 9
91 docs

Top Courses for Class 9

FAQs on Ex-9.2, Triangle And Its Angles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What are the angles in a triangle?
Ans. In a triangle, the sum of all three angles is always 180 degrees. This is known as the Angle Sum Property of a triangle.
2. How can I find the measure of an angle in a triangle if I know the measures of the other two angles?
Ans. To find the measure of an angle in a triangle, subtract the sum of the measures of the other two angles from 180 degrees. For example, if the measures of two angles in a triangle are 30 degrees and 60 degrees, then the measure of the third angle can be found by subtracting 30 + 60 = 90 from 180, resulting in 90 degrees.
3. Can a triangle have two right angles?
Ans. No, a triangle cannot have two right angles. The sum of the angles in a triangle is always 180 degrees. Since a right angle measures 90 degrees, if a triangle had two right angles, the sum of the angles would be 180 + 90 + 90 = 360 degrees, which is not possible.
4. What is the relationship between the sides and angles in a triangle?
Ans. In a triangle, the largest angle is always opposite the longest side, and the smallest angle is opposite the shortest side. This relationship can be used to determine which angle is the largest or smallest by comparing the lengths of the sides.
5. Can a triangle have all three angles greater than 90 degrees?
Ans. No, a triangle cannot have all three angles greater than 90 degrees. If all three angles were greater than 90 degrees, the sum of the angles would exceed 180 degrees, which is not possible in a triangle.
91 docs
Download as PDF
Explore Courses for Class 9 exam

Top Courses for Class 9

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Ex-9.2

,

study material

,

Ex-9.2

,

past year papers

,

Class 9

,

Objective type Questions

,

Summary

,

Class 9

,

Semester Notes

,

practice quizzes

,

Class 9

,

Viva Questions

,

Triangle And Its Angles

,

Triangle And Its Angles

,

shortcuts and tricks

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

,

Free

,

ppt

,

mock tests for examination

,

Extra Questions

,

MCQs

,

Sample Paper

,

Ex-9.2

,

Triangle And Its Angles

,

video lectures

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

,

Previous Year Questions with Solutions

,

Important questions

,

Exam

,

Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

,

pdf

;