{"id":545507,"date":"2021-10-04T11:01:30","date_gmt":"2021-10-04T11:01:30","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545507"},"modified":"2021-10-05T09:35:00","modified_gmt":"2021-10-05T09:35:00","slug":"rd-sharma-solutions-for-class-9-maths-chapter-9-triangle-and-its-angles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-9-triangle-and-its-angles\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 9\u2013Triangle and its Angles"},"content":{"rendered":"\n

Class 9: Maths Chapter 9 solutions. Complete Class 9 Maths Chapter 9 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 9\u2013Triangle and its Angles<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 9, Class 9 Maths Chapter 9 solutions<\/p>\n\n\n\n

Exercise 9.1 Page No: 9.9<\/p>\n\n\n\n

Question 1: In a \u0394ABC, if \u2220A = 550<\/sup>, \u2220B = 400<\/sup>, find \u2220C.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: \u2220A = 550<\/sup>, \u2220B = 400<\/sup><\/p>\n\n\n\n

We know, sum of all angles of a triangle is 1800<\/sup><\/p>\n\n\n\n

\u2220A + \u2220B + \u2220C = 1800<\/sup><\/p>\n\n\n\n

550 <\/sup>+ 400 <\/sup>+ \u2220C=1800<\/sup><\/p>\n\n\n\n

950 <\/sup>+ \u2220C = 1800<\/sup><\/p>\n\n\n\n

\u2220C = 1800<\/sup> \u2212 950<\/sup><\/p>\n\n\n\n

\u2220C = 850<\/sup><\/p>\n\n\n\n

Question 2: If the angles of a triangle are in the ratio 1:2:3, determine three angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Angles of a triangle are in the ratio 1:2:3 (Given)<\/p>\n\n\n\n

Let the angles be x, 2x, 3x<\/p>\n\n\n\n

Sum of all angles of triangles = 1800<\/sup><\/p>\n\n\n\n

x + 2x + 3x = 1800<\/sup><\/p>\n\n\n\n

6x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup>\/6<\/p>\n\n\n\n

x = 300<\/sup><\/p>\n\n\n\n

Answer:<\/p>\n\n\n\n

x = 300<\/sup><\/p>\n\n\n\n

2x = 2(30)0<\/sup> = 600<\/sup><\/p>\n\n\n\n

3x = 3(30) 0<\/sup> = 900<\/sup><\/p>\n\n\n\n

Question 3: The angles of a triangle are (x \u2212 40)0<\/sup>, (x \u2212 20) 0 <\/sup>and (1\/2 x \u2212 10) 0<\/sup>. Find the value of x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The angles of a triangle are (x \u2212 40)0<\/sup>, (x \u2212 20) 0 <\/sup>and (1\/2 x \u2212 10) 0<\/sup><\/p>\n\n\n\n

Sum of all angles of triangle = 1800<\/sup><\/p>\n\n\n\n

(x \u2212 40)0<\/sup> + (x \u2212 20) 0 <\/sup>+ (1\/2 x \u2212 10) 0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

5\/2 x \u2013 700<\/sup> = 1800<\/p>\n\n\n\n

5\/2 x = 1800<\/sup> + 700<\/sup><\/p>\n\n\n\n

5x = 2(250) 0<\/sup><\/p>\n\n\n\n

x = 5000<\/sup>\/5<\/p>\n\n\n\n

x = 1000<\/sup><\/p>\n\n\n\n

Question 4: The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 100<\/sup>, find the three angles.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

The difference between two consecutive angles is 100<\/sup> (given)<\/p>\n\n\n\n

Let x, x + 100<\/sup>, x + 200<\/sup> be the consecutive angles<\/p>\n\n\n\n

x + x + 100<\/sup> + x + 200<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x = 1800<\/sup>\u2013 300<\/sup><\/p>\n\n\n\n

3x = 1500<\/sup><\/p>\n\n\n\n

or x = 500<\/sup><\/p>\n\n\n\n

Again,<\/p>\n\n\n\n

x + 100<\/sup> = 500<\/sup> + 100<\/sup> = 600<\/sup><\/p>\n\n\n\n

x+200<\/sup> = 500<\/sup> + 200<\/sup> = 700<\/sup><\/p>\n\n\n\n

Answer: Three angles are 500<\/sup>,600<\/sup> and 700<\/sup>.<\/p>\n\n\n\n

Question 5: Two angles of a triangle are equal and the third angle is greater than each of those angles by 300<\/sup>. Determine all the angles of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Two angles of a triangle are equal and the third angle is greater than each of those angles by 300<\/sup>. (Given)<\/p>\n\n\n\n

Let x, x, x + 300<\/sup> be the angles of a triangle.<\/p>\n\n\n\n

Sum of all angles in a triangle = 1800<\/sup><\/p>\n\n\n\n

x + x + x + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x + 300<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x = 1500<\/sup><\/p>\n\n\n\n

or x = 500<\/sup><\/p>\n\n\n\n

And x + 300<\/sup> = 500<\/sup> + 300<\/sup> = 800<\/sup><\/p>\n\n\n\n

Answer: Three angles are 500<\/sup>, 500<\/sup> and 800<\/sup>.<\/p>\n\n\n\n

Question 6: If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

One angle of a triangle is equal to the sum of the other two angles (given)<\/p>\n\n\n\n

To Prove: One of the angles is 900<\/sup><\/p>\n\n\n\n

Let x, y and z are three angles of a triangle, where<\/p>\n\n\n\n

z = x + y \u2026(1)<\/p>\n\n\n\n

Sum of all angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

x + y + z = 1800<\/sup><\/p>\n\n\n\n

z + z = 1800<\/sup> (Using equation (1))<\/p>\n\n\n\n

2z = 1800<\/sup><\/p>\n\n\n\n

z = 900<\/sup> (Proved)<\/p>\n\n\n\n

Therefore, triangle is a right angled triangle.<\/p>\n\n\n\n

Exercise 9.2 Page No: 9.18<\/p>\n\n\n\n

Question 1: The exterior angles, obtained on producing the base of a triangle both ways are 1040<\/sup> and 1360<\/sup>. Find all the angles of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2220ACD = \u2220ABC + \u2220BAC [Exterior angle property]<\/p>\n\n\n\n

Find \u2220ABC:<\/p>\n\n\n\n

\u2220ABC + \u2220ABE = 1800 <\/sup>[Linear pair]<\/p>\n\n\n\n

\u2220ABC + 1360 <\/sup>= 1800<\/sup><\/p>\n\n\n\n

\u2220ABC = 440<\/sup><\/p>\n\n\n\n

Find \u2220ACB:<\/p>\n\n\n\n

\u2220ACB + \u2220ACD = 1800<\/sup> [Linear pair]<\/p>\n\n\n\n

\u2220ACB + 1040<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220ACB = 760<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Sum of all angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

\u2220A + 440<\/sup> + 760<\/sup> = 1800<\/sup><\/p>\n\n\n\n

\u2220A = 1800 <\/sup>\u2212 440 <\/sup>\u2212760<\/sup><\/p>\n\n\n\n

\u2220 A = 600<\/sup><\/p>\n\n\n\n

Answer: Angles of a triangle are \u2220 A = 600<\/sup>, \u2220B = 440<\/sup> and \u2220C = 760<\/sup><\/p>\n\n\n\n

Question 2: In a \u25b3ABC, the internal bisectors of \u2220B and \u2220C meet at P and the external bisectors of \u2220B and \u2220C meet at Q. Prove that \u2220BPC + \u2220BQC = 1800<\/sup>.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

In triangle ABC,<\/p>\n\n\n\n

BP and CP are internal bisector of \u2220B and \u2220C respectively<\/p>\n\n\n\n

=> External \u2220B = 180o<\/sup> \u2013 \u2220B<\/p>\n\n\n\n

BQ and CQ are external bisector of \u2220B and \u2220C respectively.<\/p>\n\n\n\n

=> External \u2220C = 180 o<\/sup> \u2013 \u2220C<\/p>\n\n\n\n

In triangle BPC,<\/p>\n\n\n\n

\u2220BPC + 1\/2\u2220B + 1\/2\u2220C = 180o<\/sup><\/p>\n\n\n\n

\u2220BPC = 180 o<\/sup> \u2013 1\/2(\u2220B + \u2220C) \u2026. (1)<\/p>\n\n\n\n

In triangle BQC,<\/p>\n\n\n\n

\u2220BQC + 1\/2(180 o<\/sup> \u2013 \u2220B) + 1\/2(180 o<\/sup> \u2013 \u2220C) = 180 o<\/sup><\/p>\n\n\n\n

\u2220BQC + 180 o<\/sup> \u2013 1\/2(\u2220B + \u2220C) = 180 o<\/sup><\/p>\n\n\n\n

\u2220BPC + \u2220BQC = 180 o<\/sup> [Using (1)]<\/p>\n\n\n\n

Hence Proved.<\/p>\n\n\n\n

Question 3: In figure, the sides BC, CA and AB of a \u25b3ABC have been produced to D, E and F respectively. If \u2220ACD = 1050<\/sup> and \u2220EAF = 450<\/sup>, find all the angles of the \u25b3ABC.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAC = \u2220EAF = 450<\/sup> [Vertically opposite angles]<\/p>\n\n\n\n

\u2220ACD = 1800<\/sup> \u2013 1050<\/sup> = 750<\/sup> [Linear pair]<\/p>\n\n\n\n

\u2220ABC = 1050<\/sup> \u2013 450<\/sup> = 600<\/sup> [Exterior angle property]<\/p>\n\n\n\n

Question 4: Compute the value of x in each of the following figures:<\/strong><\/p>\n\n\n\n

(i)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAC = 1800<\/sup> \u2013 1200<\/sup> = 600<\/sup> [Linear pair]<\/p>\n\n\n\n

\u2220ACB = 1800<\/sup> \u2013 1120<\/sup> = 680<\/sup> [Linear pair]<\/p>\n\n\n\n

Sum of all angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

x = 1800 <\/sup>\u2212 \u2220BAC \u2212 \u2220ACB<\/p>\n\n\n\n

= 1800 <\/sup>\u2212 600 <\/sup>\u2212 680 <\/sup>= 520<\/sup><\/p>\n\n\n\n

Answer: x = 520<\/sup><\/p>\n\n\n\n

(ii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220ABC = 1800<\/sup> \u2013 1200<\/sup> = 600<\/sup> [Linear pair]<\/p>\n\n\n\n

\u2220ACB = 1800<\/sup> \u2013 1100<\/sup> = 700<\/sup> [Linear pair]<\/p>\n\n\n\n

Sum of all angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

x = \u2220BAC = 1800<\/sup> \u2212 \u2220ABC \u2212 \u2220ACB<\/p>\n\n\n\n

= 1800<\/sup> \u2013 600<\/sup> \u2013 700<\/sup> = 500<\/sup><\/p>\n\n\n\n

Answer: x = 500<\/sup><\/p>\n\n\n\n

(iii)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\u2220BAE = \u2220EDC = 520<\/sup> [Alternate angles]<\/p>\n\n\n\n

Sum of all angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup> \u2013 400<\/sup> \u2013 520<\/sup> = 1800 <\/sup>\u2212 920<\/sup> = 880<\/sup><\/p>\n\n\n\n

Answer: x = 880<\/sup><\/p>\n\n\n\n

(iv)<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

CD is produced to meet AB at E.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

\u2220BEC = 1800<\/sup> \u2013 450<\/sup> \u2013 500<\/sup> = 850<\/sup> [Sum of all angles of a triangle = 1800<\/sup>]<\/p>\n\n\n\n

\u2220AEC = 1800<\/sup> \u2013 850<\/sup> = 950<\/sup> [Linear Pair]<\/p>\n\n\n\n

Now, x = 950<\/sup> + 350<\/sup> = 1300<\/sup> [Exterior angle Property]<\/p>\n\n\n\n

Answer: x = 1300<\/sup><\/p>\n\n\n\n

Question 5: In figure, AB divides \u2220DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

\u2220DAC=180\u00b0\u2212108\u00b0=72\u00b0<\/p>\n\n\n\n

\u2220BAC\/\u2220DAB\u200b=1\/3\u2220DAB=3\u2220BAC\u2220BAC+\u2220DAB=\u2220DAC=72\u00b0We can write it as\u2220BAC+3\u2220BAC=72\u00b04\u2220BAC=72\u00b0\u2220BAC=72\/4\u200b=18\u00b0So we get\u2220DAB=3\u00d718\u00b0=54\u00b0\u2220DAB=\u2220BDA=54\u00b0(AB=DB)\u2220ABD=180\u00b0\u2212(54\u00b0+54\u00b0)=180\u2212108\u00b0=72\u00b0Now,\u2220DBA=72\u00b0=\u2220BAC+x (Exterior angle)So we getx=72\u00b0\u221218\u00b0=54\u00b0<\/p>\n\n\n\n

Exercise VSAQs Page No: 9.21<\/p>\n\n\n\n

Question 1: Define a triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Triangle is a three-sided polygon that consists of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees.<\/p>\n\n\n\n

Question 2: Write the sum of the angles of an obtuse triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> The sum of angles of obtuse triangle = 180\u00b0.<\/p>\n\n\n\n

Question 3: In \u25b3ABC, if \u2220B = 60<\/strong>0<\/sup>, \u2220C = 80<\/strong>0<\/sup> and the bisectors of angles \u2220ABC and \u2220ACB meet at point O, then find the measure of \u2220BOC.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

\u2220B = 600<\/sup>, \u2220C = 800<\/sup> (given)<\/p>\n\n\n\n

As per question:<\/p>\n\n\n\n

\u2220OBC = 600<\/sup>\/2 = 300<\/sup> and<\/p>\n\n\n\n

\u2220OCB = 800<\/sup>\/2 = 400<\/sup><\/p>\n\n\n\n

In triangle BOC,<\/p>\n\n\n\n

\u2220OBC + \u2220OCB + \u2220BOC = 180 0<\/sup>[Sum of angles of a triangle = 1800<\/sup>]<\/p>\n\n\n\n

300<\/sup> + 400<\/sup> + \u2220BOC = 1800<\/sup><\/p>\n\n\n\n

\u2220BOC = 1100<\/sup><\/p>\n\n\n\n

Question 4: If the angles of a triangle are in the ratio 2:1:3, then find the measure of smallest angle.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Let angles of a triangles are 2x, x and 3x, where x is the smallest angle.<\/p>\n\n\n\n

To find: measure of x.<\/p>\n\n\n\n

As, Sum of angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

2x + x + 3x = 1800<\/sup><\/p>\n\n\n\n

6x = 1800<\/sup><\/p>\n\n\n\n

x = 300<\/sup>. Answer<\/p>\n\n\n\n

Question 5: If the angles A, B and C of \u25b3ABC satisfy the relation B \u2013 A = C \u2013 B, then find the measure of \u2220B.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Sum of angles of a triangle = 1800<\/sup><\/p>\n\n\n\n

A + B + C = 1800<\/sup> \u2026(1)<\/p>\n\n\n\n

B \u2013 A = C \u2013 B \u2026(Given)<\/p>\n\n\n\n

2B = C + A \u2026(2)<\/p>\n\n\n\n

(1) => 2B + B = 1800<\/sup><\/p>\n\n\n\n

3B =1800<\/sup><\/p>\n\n\n\n

Or B = 600<\/sup><\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 9\u2013Triangle and its Angles<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 9\u2013Triangle and its Angles PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n