RD Sharma Solutions Class 9 Triangle And Its Angles Exercise 9.2

RD Sharma Solutions Class 9 Chapter 9 Exercise 9.2

RD Sharma Class 9 Solutions Chapter 9 Ex 9.2 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:

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\(\angle ACD=\angle ABC+\angle BAC\;\;\;\;\;\;\;\;\;[Exterior\;angle\;property]\\ Now\;\angle ABC=180^{0}-136^{0}=44^{0}\;\;\;\;\;\;[Linera\;pair]\\ \angle ACB=180^{0}-104^{0}=76^{0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[Linera\;pair]\\\\ Now,\\ In\;\Delta ABC\\ \angle A+\angle ABC+\angle ACB=180^{0}\;\;\;\;\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\\ \Rightarrow \angle A+44^{0}+76^{0}=180^{0}\\ \Rightarrow \angle A=180^{0}-44^{0}-76^{0}\\ \Rightarrow \angle A=60^{0}\)

 

 

Q2) In a triangle ABC, the internal bisectors of \(\angle B\;and\;\angle C\) meet at P and the external bisectors of \(\angle B\;and\;\angle C\) meet at Q. Prove that \(\angle BPC+\angle BQC=180^{0}\).

Solution:

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\(Let\;\angle ABD=2x\;and\;\angle ACE=2y\\ \angle ABC=180^{0}-2x\;\;\;\;\;\;[Linera\;pair]\\ \angle ACB=180^{0}-2y\;\;\;\;\;\;[Linera\;pair]\\ \angle A+\angle ABC+\angle ACB=180^{0}\;\;\;\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\\ \Rightarrow \angle A+180^{0}-2x+180^{0}-2y=180^{0}\\ \Rightarrow -\angle A+2x+2y=180^{0}\\ \Rightarrow x+y=90^{0}+\frac{1}{2}\angle A\)

\(Now\;in\;\Delta BQC\\ x+y+\angle BQC=180^{0}\;\;\;\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\\ \Rightarrow 90^{0}+\frac{1}{2}\angle A+\angle BQC=180^{0}\\ \Rightarrow \angle BQC=90^{0}-\frac{1}{2}\angle A….(i)\\ and\;we\;know\;that\;\angle BPC=90^{0}+\frac{1}{2}\angle A….(ii)\\ Adding\;(i)\;and\;(ii)\;we\;get\\ \angle BPC+\angle BQC=180^{0}\)

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 \(\angle ACD=105^{0}\;and\;\angle EAF=45^{0}\), find all the angles of the triangle ABC.

Solution:

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\(\angle BAC=\angle EAF=45^{0}\;\;\;\;\;[Vertically\;opposite\;angles]\\ \angle ABC=105^{0}-45^{0}=60^{0}\;\;\;\;\;[Exterior\;angle\;property]\\ \angle ACD=180^{0}-105^{0}=75^{0}\;\;\;[Linear\;pair]\)

 

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

(i)

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Solution:

\(\angle BAC=180^{0}-120^{0}=60^{0}\;\;\;\;\;[Linear\;pair]\\ \angle ACB=180^{0}-112^{0}=68^{0}\;\;\;\;\;\;\;\;\;\;[Linear\;pair]\\ ∴ x=180^{0}-\angle BAC-\angle ACB=180^{0}-60^{0}-68^{0}=52^{0}\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\)

 

(ii)

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Solution:

\(\angle ABC=180^{0}-120^{0}=60^{0}\;\;\;\;\;[Linear\;pair]\\ \angle ACB=180^{0}-110^{0}=70^{0}\;\;\;\;\;\;\;\;\;\;[Linear\;pair]\\ ∴ e \angle BAC=x=180^{0}-\angle ABC-\angle ACB\\ =180^{0}-60^{0}-70^{0}=50^{0}\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\)

 

(iii)

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Solution:

\(\angle BAE=\angle EDC=52^{0}\;\;\;\;\;[Alternate\;angles]\\ ∴ \angle DEC=x=180^{0}-40^{0}-\angle EDC\\ =180^{0}-40^{0}-52^{0}\\ =180^{0}-92^{0}\\ =88^{0}\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\)

 

 

(iv) 

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Solution:

CD is produced to meet AB at E.

\(\angle BEC=180^{0}-45^{0}-50^{0}\\ =85^{0}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[Sum\;of\;all\;angles\;of\;a\;triangle]\\ \angle AEC=180^{0}-85^{0}=95^{0}\;\;\;\;[Linear\;pair]\\ ∴ x=95^{0}+35^{0}=130^{0}\;\;\;\;[Exterior\;angle\;property]\).

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

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Solution:

\(Let\;\angle BAD=Z,\angle BAC=3Z\\ \Rightarrow \angle BDA=\angle BAD=Z\;\;\;\;(∵ AB=DB)\\\\ Now\;\angle BAD+\angle BAC+108^{0}=180^{0}\;\;\;\;\;[Linear\;pair]\\ \Rightarrow Z+3Z+108^{0}=180^{0}\\ \Rightarrow 4Z=72^{0}\\ \Rightarrow Z=18^{0}\\\\ Now,\;In\;\Delta ADC\\ \angle ADC+\angle ACD=108^{0}\;\;\;\;[Exterior\;angle\;property]\\ \Rightarrow x+18^{0}=180^{0}\\ \Rightarrow x=90^{0}\)

 

Q6) ABC is a triangle. The bisector of the exterior angle at B and the bisector of \(\angle C\) intersect each other at D. Prove that \(\angle D=\frac{1}{2}\angle A\).

Solution:

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\(Let\;\angle ABE=2x\;and\;\angle ACB=2y\\ \angle ABC=180^{0}-2x\;\;\;\;\;[Linear\;pair]\\∴\angle A=180^{0}-\angle ABC-\angle ACB\;\;\;\;\;[Angle\;sum\;property]\\ =180^{0}-180^{0}+2x+2y\\ =2(x-y)\;\;\;\;\;\;\;…..(i)\\\\ Now,\;\angle D=180^{0}-\angle DBC-\angle DCB\\ \Rightarrow \angle D=180^{0}-(x+180^{0}-2x)-y\\ \Rightarrow \angle D=180^{0}-x-180^{0}+2x-y\\ =(x-y)\\ =\frac{1}{2}\angle A\;\;\;\;\;\;…..from(i)\\ Hence,\;\angle D=\frac{1}{2}\angle A\).

 

Q7) In figure 9.36, \(AC\perp CE\;and\;\angle A:\angle B:\angle C=3:2:1,\;find\;\)

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Solution:

\(\angle A:\angle B:\angle C=3:2:1\\\\ Let\;the\;angles\;be\;3x,\;2x\;and\;x\\ \Rightarrow 3x+2x+x=180^{0}\;\;\;\;\;\;[Angle\;sum\;property]\\ \Rightarrow 6x=180^{0}\\ \Rightarrow x=30^{0}=\angle ACB\\\\ ∴ \angle ECD=180^{0}-\angle ACB-90^{0}\;\;\;\;\;[Linear\;pair]\\ =180^{0}-30^{0}-90^{0}\\ =60^{0}\\ ∴ \angle ECD=60^{0}\)

 

 

Q8) In figure 9.37, \(AM\perp BC\;and\;AN\;is\;the\;bisector\;of\angle A.\;If\;\angle B=65^{0}\;and\;\angle C=33^{0},\;find\;\angle MAN.\).

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Solution:

\(Let\;\angle BAN=\angle NAC=x\;\;\;\;\;[∵ AN\;bisects\;\angle A]\\\\ ∴ \angle ANM=x+33^{0}\;\;\;\;\;[Exterior\;angle\;property]\\\\ In\;\Delta AMB\\ \angle BAM=90^{0}-65^{0}=25^{0}\;\;\;\;\;[Exterior\;angle\;property]\\\\ ∴ \angle MAN=\angle BAN-\angle BAM=(x-25)^{0}\\\\ Now\;in\;\Delta MAN,\\ (x-25)^{0}+(x+33)^{0}+90^{0}=180^{0}\;\;\;\;\;\;[Angle\;sum\;property]\\ \Rightarrow 2x+8^{0}=90^{0}\\ \Rightarrow 2x=82^{0}\\ \Rightarrow x=41^{0}\\\\ ∴ MAN=x-25^{0}\\ =41^{0}-25^{0}\\ =16^{0}\)

 

Q9) In a triangle ABC, AD bisects \(\angle A\;and\;\angle C>\angle B.\;Prove\;that\;\angle ADB>\angle ADC.\).

Solution:

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\(∵ \angle C>\angle B\)                               [Given]

\(\Rightarrow \angle C+x>\angle B+x\)                 [Adding x on both sides]

⇒  180° – ∠ ADC>180^{0} – ∠ ADB

⇒  -∠ ADC> – ∠ ADB

⇒  ∠ ADB > ∠ ADC

Hence proved.

 

Q10) In triangle ABC, \(BD\perp AC\;and\;CE\perp AB.\;If\;BD\;and\;CE\;intersect\;at\;O,\;prove\;that\;\angle BOC=180^{0}-\angle A.\).

Solution:

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In quadrilateral AEOD

\(\angle A+\angle AEO+\angle EOD+\angle ADO=360^{0}\\ \Rightarrow \angle A+90^{0}+90^{0}+\angle EOD=360^{0}\\ \Rightarrow \angle A+\angle BOC=180^{0}\;\;\;\;\;\;\;[∵ \angle EOD=\angle BOC\;vertically\;opposite\;angles]\\ \Rightarrow \angle BOC=180^{0}-\angle A\)

 

Q11) In figure 9.38, AE bisects \(\angle CAD\;and\;\angle B=\angle C.\;Prove\;that\;AE\parallel BC.\)

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Solution:

\(Let\;\angle B=\angle C=x\\ Then,\\ \angle CAD=\angle B+\angle C=2x\;\;\;\;\;(exterior\;angle)\\ \Rightarrow \frac{1}{2}\angle CAD=x\\ \Rightarrow \angle EAC=x\\ \Rightarrow \angle EAC=\angle C\\\)

These are alternate interior angles for the lines AE and BC

\(∴ AE\parallel BC\)

 

Q12) In figure 9.39, \(AB\parallel DE.\;Find\;\angle ACD.\)

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Solution:

Since \(AB\parallel DE\)

\(∴ \angle ABC=\angle CDE=40^{0}\;\;\;\;\;\;[Alternate\;angles]\\\\ ∴ \angle ACB=180^{0}-\angle ABC-\angle BAC\\ =180^{0}-40^{0}-30^{0}\\ =110^{0}\\\\ ∴ \angle ACD=180^{0}-110^{0}\;\;\;\;\;[Linear\;pair]\\ =70^{0}\)

 

  

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°.

(ii) A triangle can have two right angles.

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

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

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

(vi) A triangle can have two obtuse angles.

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

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

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

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

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

Solution:

(i) T

(ii) F

(iii) F

(iv) F

(v) T

(vi) F

(vii) T

(viii) T

(ix) F

(x) T

(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