RD Sharma 2020 solution class 9 chapter 11 Triangles and Its Angles Exercise 11.2

Exercise 11.2

Page-11.19

Question 1:

The exterior angles, obtained on producing the base of a triangle both way are 104° and 136°. Find all the angles of the triangle.

Answer 1:

In the given problem, the exterior angles obtained on producing the base of a triangle both ways are and . So, let us draw ΔABC and extend the base BC, such that:

Here, we need to find all the three angles of the triangle.

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

Similarly, EBC is a straight line, so we get,

Further, using angle sum property in ΔABC

Therefore,.

Question 2:

In the given figure, the sides BC, CA and AB of a Δ ABC have been produced to D, E and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the Δ ABC.

Answer 2:

In the given ΔABC, and . We need to find .

Here, are vertically opposite angles. So, using the property, “vertically opposite angles are equal”, we get,

Further, BCD is a straight line. So, using linear pair property, we get,

Now, in ΔABC, using “the angle sum property”, we get,

Therefore,.

Page-11.20

Question 3:

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

(i)

(ii)

(iii)

Answer 3:

In the given problem, we need to find the value of x

(i) In the given ΔABC, and

Now, BCD is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Similarly, EAC is a straight line. So, we get,

Further, using the angle sum property of a triangle,

In ΔABC

Therefore,

(ii) In the given ΔABC, and

Here, BCD is a straight line. So, using the property, “the angles forming a linear pair are supplementary” we get,

Similarly, EBC is a straight line. So, we get

Further, using the angle sum property of a triangle,

In ΔABC

Therefore,

(iii) In the given figure,and

      

Here,and AD is the transversal, so form a pair of alternate interior angles. Therefore, using the property, “alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔDEC

Therefore,

Question 4:

In the given figure, AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.
 

Answer 4:

In the given figure,and. We need to find the value of

 

Since, 

Let, 

Applying the angle sum property of the triangle, in ΔABC, we get,

Thus,

Further, BCD is a straight line. So, applying the property, “the angles forming a linear pair are supplementary”, we get,

Therefore,.

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Question 5:

In the given figure, AB || DE. Find ∠ACD.

Answer 5:

In the given problem,

We need to find

Now,and AE is the transversal, so using the property, “alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔDCE

Further, ACE is a straight line, so using the property, “the angles forming a linear pair are supplementary”, we get,

Therefore,.

Question 6:

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

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

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

Answer 6:

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

According to the angle sum property of the triangle

In ΔABC

Hence, the given statement is.

(ii) A triangle can have two right angles.

According to the angle sum property of the triangle

In ΔABC

Now, if there are two right angles in a triangle

Let

Then, 

(This is not possible.)

Therefore, the given statement is.

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

According to the angle sum property of the triangle

In ΔABC

Now, If all the three angles of a triangle is less than

Then, 

Therefore, the given statement is.

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

According to the angle sum property of the triangle

In ΔABC

Now, if all the three angles of a triangle is greater than

Then, 

Therefore, the given statement is.

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

According to the angle sum property of the triangle

In ΔABC

Now, if all the three angles of a triangle are equal to

Then, 

Therefore, the given statement is.

(vi) A triangle can have two obtuse angles.

According to the angle sum property of the triangle

In ΔABC

Now, if a triangle has two obtuse angles

Then, 

Therefore, the given statement is.

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

According to the angle sum property of the triangle

In ΔABC

Now, if a triangle will have more than one obtuse angle

Then, 

Therefore, the given statement is.

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

 

According to the angle sum property of the triangle

In ΔABC

Now, if it is a right angled triangle

Then, 

Also if one of the angle’s is obtuse

This is not possible.

Thus, if one angle of a triangle is obtuse, then it cannot be a right angled triangle.

Therefore, the given statement is.

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

According to the exterior angle property, an exterior angle of a triangle is equal to the sum of the two opposite interior angles.

In ΔABC

Let x be the exterior angle

So,

Now, if x is less than either of its interior opposite angles

Therefore, the given statement is.

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

According to exterior angle theorem,

Therefore, the given statement is.

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

According to exterior angle theorem,

Since, the exterior angle is the sum of its interior angles.

Thus, 

Therefore, the given statement is.

Question 7:

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.

Answer 7:

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

As we know, that according to the angle sum property, sum of all the angles of a triangle is 180°.

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

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

As according to the property: An exterior angle of a triangle is equal to the sum of two interior opposite angles. Therefore, it has to be greater than either of them.

(iv) A triangle cannot have more than one right angle.

As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one right angle the sum would exceed 180 °.

(v) A triangle cannot have more than one obtuse angle

As the sum of all the angles of a triangle is 180°. So, if the triangle has more than one obtuse angle the sum would exceed 180 °.

Question 8:

In a Δ 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 = 180°.

Answer 8:

In the given problem, BP and CP are the internal bisectors of respectively. Also, BQ and CQ are the external bisectors of respectively. Here, we need to prove:

We know that if the bisectors of anglesand of ΔABC meet at a point O then .

Thus, in ΔABC

              ……(1)

Also, using the theorem, “if the sides AB and AC of a ΔABC are produced, and the external bisectors of and meet at O, then”.

Thus, ΔABC

$\angle BQC=90°-\frac{1}{2}\angle A ......\left(2\right)$                

Adding (1) and (2), we get

Thus,

Hence proved.

Question 9:

In the given figure, compute the value of x.

Answer 9:

In the given figure,, and

Here, we will produce AD to meet BC at E

Now, using angle sum property of the triangle

In ΔAEB

Further, BEC is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Also, using the property, “an exterior angle of a triangle is equal to the sum of its two opposite interior angles”

In ΔDEC, x is its exterior angle

Thus,



Therefore,.

Page-11.22

Question 10:

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

Answer 10:

In the given figure,and

Since,and angles opposite to equal sides are equal. We get,

$\angle BDA=\angle BAD .....\left(1\right)$

Also, EAD is a straight line. So, using the property, “the angles forming a linear pair are supplementary”, we get,

Further, it is given  AB divides in the ratio 1 : 3.

So, let

$\angle DAB=y, \angle BAC=3y$

Thus, 

$$\begin{gathered} y+3y=\angle DAC \\ \Rightarrow 4y=72° \\ \Rightarrow y=\frac{72°}{4} \\ \Rightarrow y=18° \end{gathered}$$

Hence, $\angle DAB=18°, \angle BAC=3\times 18°=54°$

Using (1)

Now, in ΔABC , using the property, “exterior angle of a triangle is equal to the sum of its two opposite interior angles”, we get,

$$\begin{gathered} \angle EAC=\angle ADC+x \\ \Rightarrow 108°=18°+x \\ \Rightarrow x=90° \end{gathered}$$

Therefore,.

Question 11:

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 = $\frac{1}{2}$ ∠A.

Answer 11:

In the given ΔABC, the bisectors of and intersect at D

We need to prove:

Now, using the exterior angle theorem,

$\angle ABE=\angle BAC+\angle ACB$   .….(1)

 $\mathrm{As} \angle ABE \mathrm{and} \angle ACB \mathrm{are} \mathrm{bisected}$

$\angle DCB=\frac{1}{2}\angle ACB$

Also,

$\angle DBA=\frac{1}{2}\angle ABE$

Further, applying angle sum property of the triangle

In  ΔDCB

$$\begin{gathered} \angle CDB+\angle DCB+\angle CBD=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB+\left(\angle DBA+\angle ABC\right)=180° \end{gathered}$$

  $\angle CDB+\frac{1}{2}\angle ACB+\left(\frac{1}{2}\angle ABE+\angle ABC\right)=180° .....\left(2\right)$

Also, CBE is a straight line, So, using linear pair property

$$\begin{gathered} \Rightarrow \angle ABC+\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE+\frac{1}{2}\angle ABE=180° \\ \Rightarrow \angle ABC+\frac{1}{2}\angle ABE=180°-\frac{1}{2}\angle ABE .....\left(3\right) \end{gathered}$$

So, using (3) in (2)

$$\begin{gathered} \angle CDB+\frac{1}{2}\angle ACB+\left(180°-\frac{1}{2}\angle ABE\right)=180° \\ \Rightarrow \angle CDB+\frac{1}{2}\angle ACB-\frac{1}{2}\angle ABE=0 \\ \Rightarrow \angle CDB=\frac{1}{2}\left(\angle ABE-\angle ACB\right) \\ \Rightarrow \angle CDB=\frac{1}{2}\angle CAB \\ \Rightarrow \angle D=\frac{1}{2}\angle A \end{gathered}$$

Hence proved.

Question 12:

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Answer 12:

In the given ΔABC,, is the bisector of , and

We need to find

 

Now, using the angle sum property of the triangle

In ΔAMC, we get,

…….(1)

Similarly,

In ΔABM, we get,

…..(2)

So, adding (1) and (2)

Now, since AN is the bisector of

Thus, 

Now,

Therefore,.

Question 13:

In a Δ ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.

Answer 13:

In the given ΔABC, AD bisects and. We need to prove.

Let,

Also, 

As AD bisects, 

…..(1) 

Now, in ΔABD, using exterior angle theorem, we get,

Similarly,

[using (1)]

Further, it is given,

Adding to both the sides

Thus,

Hence proved.

Question 14:

In Δ ABC, BD⊥ AC and CE ⊥ AB. If BD and CE intersect at O, prove that ∠BOC = 180° − A.

Answer 14:

In the given ΔABC,and .

We need prove

Here, in ΔBDC, using the exterior angle theorem, we get,

Similarly, in ΔEBC, we get,

Adding (1) and (2), we get,

Now, on using angle sum property, 

In ΔABC, we get,

This can be written as,

Similarly, using angle sum property in ΔOBC, we get,

This can be written as,

Now, using the values of (4) and (5) in (3), we get,

Therefore,.

Hence proved

Question 15:

In the given figure, AE bisects ∠CAD and ∠B= ∠C. Prove that AE || BC.
 

Answer 15:

In the given problem, AE bisectsand

We need to prove

As,is bisected by AE

=2=2          ..........(1)

Now, using the property, “an exterior angle of a triangle in equal to the sum of the two opposite interior angles”, we get,

()

(using 1)

Hence, using the property, if alternate interior angles are equal, then the two lines are parallel, we get,

Thus,

Hence proved.