RD Sharma Solutions for Class 9 Maths Chapter 9 Exercise 9.2 are available here. All questions in Chapter 9 are prepared by subject experts to help students clear their doubts. RD Sharma Solutions Class 9 Exercise 9.2 are based on the concepts of exterior angles of a triangle, which states that if a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.
RD Sharma Solutions for Class 9 Maths Chapter 9 Triangle and Its Angles Exercise 9.2
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 9 Triangle and Its Angles Exercise 9.2 Page Number 9.18
Question 1: 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:
∠ACD = ∠ABC + ∠BAC [Exterior angle property]
Find ∠ABC:
∠ABC + ∠ABE = 1800 [Linear pair]
∠ABC + 1360 = 1800
∠ABC = 440
Find ∠ACB:
∠ACB + ∠ACD = 1800 [Linear pair]
∠ACB + 1040 = 1800
∠ACB = 760
Now,
Sum of all angles of a triangle = 1800
∠A + 440 + 760 = 1800
∠A = 1800 − 440 −760
∠ A = 600
Answer: Angles of a triangle are ∠ A = 600, ∠B = 440 and ∠C = 760
Question 2: In a △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:
In triangle ABC,
BP and CP are internal bisector of ∠B and ∠C respectively
=> External ∠B = 180o – ∠B
BQ and CQ are external bisector of ∠B and ∠C respectively.
=> External ∠C = 180 o – ∠C
In triangle BPC,
∠BPC + 1/2∠B + 1/2∠C = 180o
∠BPC = 180 o – 1/2(∠B + ∠C) …. (1)
In triangle BQC,
∠BQC + 1/2(180 o – ∠B) + 1/2(180 o – ∠C) = 180 o
∠BQC + 180 o – 1/2(∠B + ∠C) = 180 o
∠BPC + ∠BQC = 180 o [Using (1)]
Hence Proved.
Question 3: In figure, the sides BC, CA and AB of a △ABC have been produced to D, E and F respectively. If ∠ACD = 1050 and ∠EAF = 450, find all the angles of the △ABC.
Solution:
∠BAC = ∠EAF = 450 [Vertically opposite angles]
∠ACD = 1800 – 1050 = 750 [Linear pair]
∠ABC = 1050 – 450 = 600 [Exterior angle property]
Question 4: Compute the value of x in each of the following figures:
(i)
Solution:
∠BAC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1120 = 680 [Linear pair]
Sum of all angles of a triangle = 1800
x = 1800 − ∠BAC − ∠ACB
= 1800 − 600 − 680 = 520
Answer: x = 520
(ii)
Solution:
∠ABC = 1800 – 1200 = 600 [Linear pair]
∠ACB = 1800 – 1100 = 700 [Linear pair]
Sum of all angles of a triangle = 1800
x = ∠BAC = 1800 − ∠ABC − ∠ACB
= 1800 – 600 – 700 = 500
Answer: x = 500
(iii)
Solution:
∠BAE = ∠EDC = 520 [Alternate angles]
Sum of all angles of a triangle = 1800
x = 1800 – 400 – 520 = 1800 − 920 = 880
Answer: x = 880
(iv)
Solution:
CD is produced to meet AB at E.
∠BEC = 1800 – 450 – 500 = 850 [Sum of all angles of a triangle = 1800]
∠AEC = 1800 – 850 = 950 [Linear Pair]
Now, x = 950 + 350 = 1300 [Exterior angle Property]
Answer: x = 1300
Question 5: In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.
Solution:
RD Sharma Solutions for Class 9 Maths Chapter 9 Triangle and Its Angles Exercise 9.2
RD Sharma Solutions for Class 9 Maths Chapter 9 Triangle and Its Angles Exercise 9.2 are based on the exterior angles of a triangle. If the side BC of a triangle ABC is produced to form ray BD, then ∠ACD is called an exterior angle of triangle ABC. Students are advised to practise all the questions to score high marks in the final examination.
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