RD Sharma 2020 2021 Solutions for Class 9 Maths Chapter 11 Triangle And Its Angles are provided here with simple step-by-step explanations. These solutions for Triangle And Its Angles are extremely popular among class 9 students for Maths Triangle And Its Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2020 2021 Book of class 9 Maths Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2020 2021 Solutions. All RD Sharma 2020 2021 Solutions for class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 11.10:

#### Question 1:

In a Δ *ABC, *if ∠*A* = 55°, ∠*B* = 40°, find ∠*C*.

#### Answer:

$\angle A+\angle B+\angle C=180\xb0\left[\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0.\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 55\xb0+40\xb0+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 95\xb0+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C=180\xb0-95\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle C=85\xb0$

#### Page No 11.10:

#### Question 2:

If the angles of a triangle are in the ratio 1 : 2 : 3, determine three angles.

#### Answer:

Let the angles of the given triangle be of *x*º, 2*x*º and 3*x*º. Then,

$\therefore x+2x+3x=180\left[\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 6x=180\phantom{\rule{0ex}{0ex}}\Rightarrow x=30\phantom{\rule{0ex}{0ex}}$

Hence, the angles of the triangle are 30º, 60º and 90º.#### Page No 11.10:

#### Question 3:

The angles of a triangle are (*x* − 40)°, (*x* − 20)° and ${\left(\frac{1}{2}x-10\right)}^{\xb0}$. Find the value of *x*.

#### Answer:

Given angles are

$\therefore \left(x-40\right)+\left(x-20\right)+\left(\frac{1}{2}x-10\right)=180\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{2}x=180+70\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{2}x=250\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{250\times 2}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x=100$

Hence, the value of *x* is 100°.

#### Page No 11.10:

#### Question 4:

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30°. Determine all the angles of the triangle.

#### Answer:

Let the two equal angles are *x*°, then the third angle will be (*x* + 30)°.

Therefore, the angles of the given triangle are 50°, 50° and 80°.

#### Page No 11.10:

#### Question 5:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.

#### Answer:

Let ABC be a triangle such that

$\angle A=\angle B+\angle C\left[\mathrm{Since},\mathrm{one}\mathrm{angle}\mathrm{is}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{two}\right]\phantom{\rule{0ex}{0ex}}\therefore \angle A+\angle B+\angle C=180\xb0\left[\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle A+\angle A=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle A=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle A=90\xb0$

#### Page No 11.10:

#### Question 6:

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60°?

(v) All angles less than 60°?

(vi) All angles equal to 60°?

#### Answer:

(i) Let a triangle ABC has two angles equal to . We know that sum of the three angles of a triangle is 180°.

Hence, if two angles are equal to , then the third one will be equal to zero which implies that A, B, C is collinear, or we can say ABC is not a triangle

A triangle can’t have two right angles.

(ii) Let a triangle ABC has two obtuse angles

This implies that sum of only two angles will be equal to more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can’t have two obtuse angles.

(iii) Let a triangle ABC has two acute angles.

This implies that sum of two angles will be less than . Hence third angle will be the difference of 180° and sum of both acute angles

Therefore, a triangle can have two acute angles.

(iv) Let a triangle ABC having angles are more than 60°.

This implies that the sum of three angles will be more than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can’t have all angles more than .

(v) Let a triangle ABC having anglesare less than 60°.

This implies that the sum of three angles will be less than 180° which contradicts the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can’t have all angles less than 60°.

(vi) Let a triangle ABC having angles all equal to 60°.

This implies that the sum of three angles will be equal to 180° which satisfies the theorem sum of all angles in a triangle is always equals 180°.

Therefore, a triangle can have all angles equal to 60°.

#### Page No 11.10:

#### Question 7:

The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10°, find the three angles.

#### Answer:

Let the angles of a triangle are [Since, the difference between two consceutive angles is 10°]

$\therefore x+\left(x+10\right)+\left(x+20\right)=180\left[\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 3x+30=180\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=150\phantom{\rule{0ex}{0ex}}\Rightarrow x=50\phantom{\rule{0ex}{0ex}}$

Therefore, the angles of the given triangle are 50°, (50 + 10)° and (50 + 20)° i.e. 50°, 60° and 70°.

#### Page No 11.10:

#### Question 8:

*ABC* is a triangle in which ∠*A* = 72°, the internal bisectors of angles *B* and *C* meet in *O*. Find the magnitude of ∠*BOC*.

#### Answer:

Since OB and OC are the angle bisector of $\angle B\mathrm{and}\angle C$

$\angle A+\angle B+\angle C=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 72\xb0+2\angle OBC+2\angle OCB=180\xb0\left[\mathrm{Sum}\mathrm{of}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(\angle OBC+\angle OCB\right)=108\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle OBC+\angle OCB=54\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 180\xb0-\angle BOC=54\xb0\left[\mathrm{Since},\angle OBC+\angle OCB+\angle BOC=180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle BOC=126\xb0\phantom{\rule{0ex}{0ex}}$

Hence magnitude of $\angle BOC\mathrm{is}126\xb0.$

#### Page No 11.10:

#### Question 9:

The bisectors of base angles of a triangle cannot enclose a right angle in any case.

#### Answer:

Let ABC be a triangle and * BO* and* CO* be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠*ABC* and ∠*ACB* of a triangle *ABC* meet at a point *O*, then

$\angle BOC=90\xb0+\frac{1}{2}\angle A$

From the above relation it is very clear that if is equals 90° then must be equal to zero.

Now, if possible let is equals zero but on other hand it represents that *A*,* B*, *C* will be collinear, that is they do not form a triangle.

It leads to a contradiction.

Hence, the bisectors of base angles of a triangle cannot enclose a right angle in any case.

#### Page No 11.10:

#### Question 10:

If the bisectors of the base angles of a triangle enclose an angle of 135°, prove that the triangle is a right triangle.

#### Answer:

Let *ABC* be a triangle and Let *BO *and *CO* be the bisectors of the base anglerespectively.

We know that if the bisectors of angles ∠*ABC *and ∠*ACB* of a triangle *ABC *meet at a point *O*, then

$\angle BOC=90\xb0+\frac{1}{2}\angle A$

Hence the triangle is a right angled triangle.

#### Page No 11.10:

#### Question 11:

In a Δ *ABC*, ∠*ABC* = ∠*ACB* and the bisectors of ∠*ABC* and ∠*ACB* intersect at *O* such that ∠*BOC* = 120°. Show that ∠*A* =∠*B* =∠*C* = 60°.

#### Answer:

Let *ABC* be a triangle and *BO* and *CO* be the bisectors of the base anglerespectively.

*ABC*and ∠

*ACB*of a triangle

*ABC*meet at a point

*O*, then

$\angle BOC=90\xb0+\frac{1}{2}\angle A$

$\therefore 120\xb0=90\xb0+\frac{1}{2}\angle A\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0=\frac{1}{2}\angle A\phantom{\rule{0ex}{0ex}}\Rightarrow \angle A=60\xb0$

are equal as it is given that .

$\angle A+\angle B+\angle C=180\xb0\left[\mathrm{Sum}\mathrm{of}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180\xb0\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+2\angle B=180\xb0\left[\because \angle ABC=\angle ACB\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle B=60\xb0\phantom{\rule{0ex}{0ex}}$

Hence, .

#### Page No 11.10:

#### Question 12:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute angled.

#### Answer:

Let a triangle ABC having angles.

It is given that the sum of two angles are less than third one.

We know that the sum of all angles of a triangle equal to 180°.

Similarly we can prove that

Since, all angles are less than 90°.

Hence, triangle is acute angled.

#### Page No 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:

In the given problem, the exterior angles obtained on producing the base of a triangle both ways are and . So, let us draw Δ*ABC** *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 Δ*ABC*

Therefore,.

#### Page No 11.19:

#### 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:

In the given Δ*ABC*, 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 Δ*ABC*, using “the angle sum property”, we get,

Therefore,.

#### Page No 11.20:

#### Question 3:

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

(i)

(ii)

(iii)

#### Answer:

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

(i) In the given Δ*ABC*, 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 Δ*ABC*

Therefore,

(ii) In the given Δ*ABC*, 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 Δ*ABC*

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 Δ*DEC*

Therefore,

#### Page No 11.20:

#### Question 4:

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

#### Answer:

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

Since,

Let,

Applying the angle sum property of the triangle, in Δ*ABC*, we get,

Thus,

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

Therefore,.

#### Page No 11.21:

#### Question 5:

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

#### Answer:

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 Δ*DCE*

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

Therefore,.

#### Page No 11.21:

#### 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:

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

According to the angle sum property of the triangle

In Δ*ABC*

Hence, the given statement is.

(ii) A triangle can have two right angles.

According to the angle sum property of the triangle

In Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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 Δ*ABC*

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.

#### Page No 11.21:

#### 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:

(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 °.

#### Page No 11.21:

#### 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:

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 Δ*ABC *meet at a point *O *then .

Thus, in Δ*ABC*

……(1)

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

Thus, Δ*ABC*

$\angle BQC=90\xb0-\frac{1}{2}\angle A......\left(2\right)$

Adding (1) and (2), we get

Thus,

Hence proved.

#### Page No 11.21:

#### Question 9:

In the given figure, compute the value of *x.*

#### Answer:

In the given figure,, and

Here, we will produce *AD* to meet *BC* at *E*

Now, using angle sum property of the triangle

In Δ*AEB*

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 Δ*DEC*, *x* is its exterior angle

Thus,

Therefore,.

#### Page No 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:

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,

$y+3y=\angle DAC\phantom{\rule{0ex}{0ex}}\Rightarrow 4y=72\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{72\xb0}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=18\xb0$

Hence, $\angle DAB=18\xb0,\angle BAC=3\times 18\xb0=54\xb0$

Using (1)

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

$\angle EAC=\angle ADC+x\phantom{\rule{0ex}{0ex}}\Rightarrow 108\xb0=18\xb0+x\phantom{\rule{0ex}{0ex}}\Rightarrow x=90\xb0$

Therefore,.

#### Page No 11.22:

#### 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:

In the given Δ*ABC*, 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 Δ*DCB*

$\angle CDB+\angle DCB+\angle CBD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CDB+\frac{1}{2}\angle ACB+\left(\angle DBA+\angle ABC\right)=180\xb0$

$\angle CDB+\frac{1}{2}\angle ACB+\left(\frac{1}{2}\angle ABE+\angle ABC\right)=180\xb0.....\left(2\right)$

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

$\Rightarrow \angle ABC+\angle ABE=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC+\frac{1}{2}\angle ABE+\frac{1}{2}\angle ABE=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle ABC+\frac{1}{2}\angle ABE=180\xb0-\frac{1}{2}\angle ABE.....\left(3\right)\phantom{\rule{0ex}{0ex}}$

So, using (3) in (2)

$\angle CDB+\frac{1}{2}\angle ACB+\left(180\xb0-\frac{1}{2}\angle ABE\right)=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CDB+\frac{1}{2}\angle ACB-\frac{1}{2}\angle ABE=0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CDB=\frac{1}{2}\left(\angle ABE-\angle ACB\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle CDB=\frac{1}{2}\angle CAB\phantom{\rule{0ex}{0ex}}\Rightarrow \angle D=\frac{1}{2}\angle A$

Hence proved.

#### Page No 11.22:

#### Question 12:

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

#### Answer:

In the given Δ*ABC*,, is the bisector of , and

We need to find

Now, using the angle sum property of the triangle

In Δ*AMC*, we get,

…….(1)

Similarly,

In Δ*ABM*, we get,

…..(2)

So, adding (1) and (2)

Now, since *AN *is the bisector of

Thus,

Now,

Therefore,.

#### Page No 11.22:

#### Question 13:

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

#### Answer:

In the given Δ*ABC*, *AD* bisects and. We need to prove.

Let,

Also,

As *AD* bisects,

…..(1)

Now, in Δ*AB*D, using exterior angle theorem, we get,

Similarly,

[using (1)]

Further, it is given,

Adding to both the sides

Thus,

Hence proved.

#### Page No 11.22:

#### Question 14:

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

#### Answer:

In the given Δ*ABC*,and .

We need prove

Here, in Δ*BDC*, using the exterior angle theorem, we get,

Similarly, in Δ*EBC*, we get,

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

Now, on using angle sum property,

In Δ*ABC*, we get,

This can be written as,

Similarly, using angle sum property in Δ*OBC*, we get,

This can be written as,

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

Therefore,.

Hence proved

#### Page No 11.22:

#### Question 15:

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

#### Answer:

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.

#### Page No 11.23:

#### Question 1:

Mark the correct alternative in each of the following:

If all the three angles of a triangle are equal, then each one of them is equal to

(a) 90°

(b) 45°

(c) 60°

(d) 30°

#### Answer:

In a given Δ*ABC* we are given that the three angles are equal. So,

According to the angle sum property of a triangle, in Δ*ABC*

Therefore, all the three angles of the triangle are equal to

So, the correct option is (c).

#### Page No 11.23:

#### Question 2:

If two acute angles of a right triangle are equal, then each acute is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### Answer:

In the given problem, we have a right angled triangle and the other two angles are equal.

So, In Δ*ABC*

Now, using the angle sum property of the triangle, in Δ*ABC*, we get,

()

Therefore, the correct option is (b).

#### Page No 11.23:

#### Question 3:

An exterior angle of a triangle is equal to 100° and two interior opposite angles are equal. Each of these angles is equal to

(a) 75°

(b) 80°

(c) 40°

(d) 50°

#### Answer:

In the Δ*ABC*, *CD* is the ray extended from the vertex C of Δ*ABC*. It is given that the exterior angle of the triangle is and two of the interior opposite angles are equal.

So, and_{.}

So, now using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get.

In Δ*ABC*

Therefore, each of the two opposite interior angles is

So, the correct option is (d).

#### Page No 11.23:

#### Question 4:

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

#### Answer:

In the given problem, one angle of a triangle is equal to the sum of the other two angles.

Thus,

..........(1)

Now, according to the angle sum property of the triangle

In Δ*ABC*

.........(2)

Further, using (2) in (1),

Thus,

Therefore, the correct option is (d).

#### Page No 11.23:

#### Question 5:

Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = $\frac{1}{2}$ ∠A is equal to

(a) 80°

(b) 75°

(c) 60°

(d) 90°

#### Answer:

In the given problem, side *BC* of Δ*ABC* has been produced to a point D. Such that and. Here, we need to find

Given

We get,

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

In Δ*ABC*

Also, (Using 1)

Thus,

Therefore, the correct option is (a).

#### Page No 11.24:

#### Question 6:

In Δ*ABC*, ∠*B* = ∠*C* and ray *AX* bisects the exterior angle ∠*DAC*. If ∠*DAX* = 70°, then ∠*ACB* =

(a) 35°

(b) 90°

(c) 70°

(d) 55°

#### Answer:

In the given Δ*ABC*, . D is the ray extended from point A. AX bisectsand

Here, we need to find

As ray *AX* bisects

Thus,

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

Thus,

Therefore, the correct option is (c).

#### Page No 11.24:

#### Question 7:

In a triangle, an exterior angle at a vertex is 95° and its one of the interior opposite angle is 55°, then the measure of the other interior angle is

(a) 55°

(b) 85°

(c) 40°

(d) 9.0°

#### Answer:

In the given Δ*ABC*, and

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

So,

Therefore, the correct option is (c).

#### Page No 11.24:

#### Question 8:

If the sides of a triangle are produced in order, then the sum of the three exterior angles so formed is

(a) 90°

(b) 180°

(c) 270°

(d) 360°

#### Answer:

In the given Δ*ABC*, all the three sides of the triangle are produced. We need to find the sum of the three exterior angles so produced.

Now, according to the angle sum property of the triangle

.......(1)

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

......(2)

Similarly,

.......(3)

Also,

.......(4)

Adding (2) (3) and (4)

We get,

Thus,

Therefore, the correct option is (d).

#### Page No 11.24:

#### Question 9:

In ΔABC, if ∠A = 100°, AD bisects ∠A and AD ⊥ BC. Then, ∠B =

(a) 50°

(b) 90°

(c) 40°

(d) 100°

#### Answer:

In the given Δ*ABC*,, AD bisects and .

Here, we need to find.

As, *AD* bisects,

We get,

Now, according to angle sum property of the triangle

In Δ*ABD*

Hence,

Therefore, the correct option is (c).

#### Page No 11.24:

#### Question 10:

An exterior angle of a triangle is 108° and its interior opposite angles are in the ratio 4 : 5. The angles of the triangle are

(a) 48°, 60°, 72°

(b) 50°, 60°, 70°

(c) 52°, 56°, 72°

(d) 42°, 60°, 76°

#### Answer:

In the given Δ*ABC*, an exterior angle and its interior opposite angles are in the ratio 4:5.

Let us take,

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

We get,

Thus,

Also, using angle sum property in Δ*ABC*

Thus,

Therefore, the correct option is (a).

#### Page No 11.24:

#### Question 11:

In a Δ*ABC*, if ∠*A* = 60°, ∠*B* = 80° and the bisectors of ∠*B* and ∠*C* meet at *O*, then ∠*BOC* =

(a) 60°

(b) 120°

(c) 150°

(d) 30°

#### Answer:

In the given Δ*ABC*,and . Bisectors of and meet at *O*.

We need to find

Since, *OB* is the bisector of.

Thus,

Now, using the angle sum property of the triangle

In Δ*ABC*, we get,

Similarly, in Δ*BOC*

Hence,

Therefore, the correct option is (b).

#### Page No 11.24:

#### Question 12:

Line segments *AB* and *CD* intersect at *O* such that *AC* || *DB*. If ∠*CAB* = 45° and ∠*CDB* = 55°, then ∠*BOD* =

(a) 100°

(b) 80°

(c) 90°

(d) 135°

#### Answer:

In the given problem, line segment *AB* and *CD* intersect at *O*, such that,and .

We need to find

As

Applying the property, “alternate interior angles are equal”, we get,

.......(1)

Now, using the angle sum property of the triangle

In Δ*ODB*, we get,

(using 1)

Thus,

Therefore, the correct option is (b).

#### Page No 11.24:

#### Question 13:

In the given figure, if *EC* || AB, ∠*ECD* = 70° and ∠*BDO* = 20°, then ∠*OBD* is

(a) 20°

(b) 50°

(c) 60°

(d) 70°

#### Answer:

In the given figure,,and **. **We need to find.

Here, and *CD* is the transversal, so using the property, “corresponding angles are equal”, we get

Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in Δ*OBD*, we get,

Thus,

Therefore, the correct option is (b).

#### Page No 11.24:

#### Question 14:

In the given figure, *x* +* y* =

(a) 270

(b) 230

(c) 210

(d) 190°

#### Answer:

In the given figure, we need to find

Here, *AB* and *CD *are straight lines intersecting at point *O*, so using the property, “vertically opposite angles are equal”, we get,

Further, applying the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in ΔA*OC*, we get,

Similarly, in Δ*BOD*

Thus,

Therefore, the correct option is (b).

#### Page No 11.25:

#### Question 15:

If the measures of angles of a triangle are in the ratio of 3 : 4 : 5, what is the measure of the smallest angle of the triangle?

(a) 25°

(b) 30°

(c) 45

(d) 60°

#### Answer:

In the given figure, measures of the angles of Δ*ABC** *are in the ratio. We need to find the measure of the smallest angle of the triangle.

Let us take,

Now, applying angle sum property of the triangle in Δ*ABC*, we get,

Substituting the value of *x *in,and

Since, the measure of is the smallest

Thus, the measure of the smallest angle of the triangle is

Therefore, the correct option is (c).

#### Page No 11.25:

#### Question 16:

In the given figure, if *AB* ⊥ *BC*. then *x* =

(a) 18

(b) 22

(c) 25

(d) 32

#### Answer:

In the given figure,

We need to find the value of *x*.

Now, since *AB *and *CD *are straight lines intersecting at point *O*, using the property, “vertically opposite angles are equal”, we get,

Further, applying angle sum property of the triangle

In Δ*BOC*

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

In Δ*EOC*

Further solving for *x, *we get,

Thus,

Therefore, the correct option is (b).

#### Page No 11.25:

#### Question 17:

In the given figure, what is z in terms of x and y?

(a) x + y + 180

(b) x + y − 180

(c) 180° − (x + y)

(d) x + y + 360°

#### Answer:

In the given Δ*ABC*, we need to convert *z* in terms of *x* and *y*

Now,* BC* is a straight line, so using the property, “angles forming a linear pair are supplementary”

Similarly,

Also, using the property, “vertically opposite angles are equal”, we get,

Further, using angle sum property of the triangle

Thus,

Therefore, the correct option is (b).

#### Page No 11.25:

#### Question 18:

In the given figure, for which value of *x* is *l*_{1} || *l*_{2}?

(a) 37

(b) 43

(c) 45

(d) 47

#### Answer:

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

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

Further, applying angle sum property of the triangle

In Δ*ABC*

Thus,

Therefore, the correct option is (d).

#### Page No 11.25:

#### Question 19:

In the given figure, what is *y* in terms of *x*?

(a) $\frac{3}{2}x$

(b) $\frac{4}{3}x$

(c) *x*

(d) $\frac{3}{4}x$

#### Answer:

In the given figure, we need to find *y *in terms of *x*

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

In Δ*ABC*

..........(1)

Similarly, in Δ*OCD*

(using 1)

Thus,

Therefore, the correct option is (a).

#### Page No 11.26:

#### Question 20:

In the given figure, what is the value of *x*?

(a) 35

(b) 45

(c) 50

(d) 60

#### Answer:

In the given figure, we need to find the value of *x*.

Here, *DBA *is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Now, applying the value of *y* inand

Also,

Further, applying angle sum property of the triangle

In Δ*ABC*

Thus,

Therefore, the correct option is (d).

#### Page No 11.26:

#### Question 21:

In the given figure, the value of *x* is

(a) 65°

(b) 80°

(c) 95°

(d) 120°

#### Answer:

In the given figure, we need to find the value of *x*

Here, according to the angle sum property of the triangle

In Δ*ABD*

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

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

Thus,

Therefore, the correct option is (d).

#### Page No 11.26:

#### Question 22:

In the given figure, if *BP* || *CQ* and *AC* = *BC*, then the measure of *x* is

(a) 20°

(b) 25°

(c) 30°

(d) 35°

#### Answer:

In the given figure,and

We need to find the measure of *x*

Here, we draw a line *RS* parallel to *BP*, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

Now, and *BA* is the transversal, so using the property, “alternate interior angles are equal”

Similarly, and *AC* is the transversal

........(2)

Adding (1) and (2), we get

Also, as

Using the property,”angles opposite to equal sides are equal”, we get

Further, using the property, “an exterior angle is equal to the sum of the two opposite interior angles”

In Δ*ABC*

Thus,

Therefore, the correct option is (c).

#### Page No 11.27:

#### Question 23:

In the given figure, *AB* and *CD* are parallel lines and transversal *EF* intersects them at *P* and *Q *respectively. If ∠*APR** *= 25°, ∠*RQC* = 30° and ∠*CQF* = 65°, then

(a) *x* = 55°, *y* = 40°

(b) *x* = 50°, *y* = 45°

(c) *x* = 60°, *y* = 35°

(d) *x* = 35°, *y* = 60°

#### Answer:

In the given figure,,,and

We need to find the value of *x* and *y*

Here, we draw a line *ST* parallel to *AB*, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

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

Similarly,and *EF* is the transversal

.......(2)

Adding (1) and (2), we get

Further,*FPE* is a straight line

Applying the property, angles forming a linear pair are supplementary

Also, applying angle sum property of the triangle

In Δ*PRQ*

Thus,

Therefore, the correct option is (a).

#### Page No 11.27:

#### Question 24:

The base *BC *of triangle *ABC* is produced both ways and the measure of exterior angles formed are 94° and 126°. Then, ∠*BAC* =

(a) 94°

(b) 54°

(c) 40°

(d) 44°

#### Answer:

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

Here, we need to find

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

Similarly, *EBS *is a straight line, so we get,

Further, using angle sum property in Δ*ABC*

Thus,

Therefore, the correct option is (c).

#### Page No 11.27:

#### Question 25:

If the bisectors of the acute angles of a right triangle meet at *O*, then the angle at *O* between the two bisectors is

(a) 45°

(b) 95°

(c) 135°

(d) 90°

#### Answer:

In the given problem, bisectors of the acute angles of a right angled triangle meet at *O. *We need to find .

Now, using the angle sum property of a triangle

In Δ*ABC*

Now, further multiplying each of the term by in (1)

Also, applying angle sum property of a triangle

In Δ*AOC*

Thus,

Therefore, the correct option is (c).

#### Page No 11.27:

#### Question 26:

The bisects of exterior angle at B and C of ΔABC meet at O. If ∠A = x°, then ∠BOC =

(a) $90\xb0+\frac{x\xb0}{2}$

(b) $90\xb0-\frac{x\xb0}{2}$

(c) $180\xb0+\frac{x\xb0}{2}$

(d) $180\xb0-\frac{x\xb0}{2}$

#### Answer:

In the given figure, bisects of exterior anglesand meet at *O* and

We need to find

Now, according to the theorem, “if the sides *AB* and *AC* of a Δ*ABC** *are produced to *P* and *Q* respectively and the bisectors of and intersect at *O*, therefore, we get,

Hence, in Δ*ABC*

Thus,

Therefore, the correct option is (b).

#### Page No 11.27:

#### Question 27:

In a Δ*ABC*, ∠*A* = 50° and *BC* is produced to a point *D*. If the bisectors of ∠*ABC* and ∠*ACD* meet at *E*, then ∠*E* =

(a) 25°

(b) 50°

(c) 100°

(d) 75°

#### Answer:

In the given figure, bisectors of and meet at *E* and

We need to find

Here, using the property, “an exterior angle of the triangle is equal to the sum of the opposite interior angles”, we get,

In Δ*ABC *with as its exterior angle

........(1)

Similarly, in with as its exterior angle

(*CE *and *BE* are the bisectors of and)

.......(2)

Now, multiplying both sides of (1) by

We get,

......(3)

From (2) and (3) we get,

Thus,

Therefore, the correct option is (a).

#### Page No 11.27:

#### Question 28:

The side BC of Δ*ABC* is produced to a point *D*. The bisector of ∠*A* meets side *BC* in L. If ∠*ABC** *= 30° and ∠*ACD* = 115°, then ∠*ALC* =

(a) 85°

(b) $72\frac{1}{2}\xb0$

(c) 145°

(d) none of these

#### Answer:

In the given problem, *BC* of Δ*ABC** *is produced to point *D*. bisectors of meet side* BC* at *L*, and

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

In Δ*ABC*

Now, as *AL* is the bisector of

Also, is the exterior angle of Δ*ALC*

Thus,

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

In

Thus,

Therefore the correct option is (b).

#### Page No 11.27:

#### Question 29:

In the given figure, if *l*_{1} || *l*_{2}, the value of *x* is

(a) $22\frac{1}{2}$

(b) 30

(c) 45

(d) 60

#### Answer:

In the given problem,

We need to find the value of *x*

Here, as, using the property, “consecutive interior angles are supplementary”, we get

..........(1)

Further, applying angle sum property of the triangle

In Δ*ABC*

(using 1)

Now, *AB *is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Thus,

Therefore, the correct option is (c).

#### Page No 11.28:

#### Question 30:

In Δ*RST* (See figure), what is the value of *x*?

(a) 40

(b) 90°

(c) 80°

(d) 100

#### Answer:

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

Here, according to the corollary, “if bisectors of and of a Δ*ABC* meet at a point *O*, then

In Δ*RST*

Further solving for *x*, we get,

Thus,

Therefore, the correct option is (d).

#### Page No 11.28:

#### Question 1:

The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is a/an ___________ triangle.

#### Answer:

Let the measure of three angles of the triangle be 5*x*, 3*x* and 7*x*.

Now,

$5x+3x+7x=180\xb0$ (Angle sum property of triangle)

$\Rightarrow 15x=180\xb0$

$\Rightarrow x=\frac{180\xb0}{15}=12\xb0$

$\therefore 5x=5\times 12\xb0=60\xb0,3x=3\times 12\xb0=36\xb0$ and $7x=7\times 12\xb0=84\xb0$

So, the measure of the angles of the triangle are 60º, 36º and 84º.

A triangle, each of whose angles is acute, is called an acute triangle. Thus, the triangle is an acute triangle.

The angles of a triangle are in the ratio 5 : 3 : 7. The triangle is a/an ____acute____ triangle.

#### Page No 11.28:

#### Question 2:

Angles of a triangle are in the ratio 2 : 4 : 3. The measure of the smallest angle of the triangle is ________.

#### Answer:

Let the measure of three angles of the triangle be 2*x*, 4*x* and 3*x*.

Now,

$2x+4x+3x=180\xb0$ (Angle sum property of triangle)

$\Rightarrow 9x=180\xb0$

$\Rightarrow x=\frac{180\xb0}{9}=20\xb0$

$\therefore 2x=2\times 20\xb0=40\xb0,4x=4\times 20\xb0=80\xb0$ and $3x=3\times 20\xb0=60\xb0$

So, the measure of the angles of the triangle are 40º, 80º and 60º.

Thus, the measure of the smallest angle of the triangle is 40º.

Angles of a triangle are in the ratio 2 : 4 : 3. The measure of the smallest angle of the triangle is _____40º_____.

#### Page No 11.28:

#### Question 3:

The number of triangles that can be drawn the measure of whose angles are 53°, 64° and 63°, is _________.

#### Answer:

We know that the sum of the angles of a triangle is 180º.

The given angles are 53°, 64° and 63°.

Sum of the given angles = 53° + 64° + 63° = 180º

Here, the sum of the angles of the triangle is 180º. But, the measure of sides of the triangles is not known. So, infinitely many triangles can be drawn and sum of the angles of every triangle is 180º.

The number of triangles that can be drawn the measure of whose angles are 53°, 64° and 63°, is ____infinite many____.

#### Page No 11.28:

#### Question 4:

If the measure of one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles is _________.

#### Answer:

In âˆ†ABC, ∠A = 130º.

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2}.....\left(1\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}.....\left(2\right)$

Now,

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0$ (Angle sum property of triangle)

$\Rightarrow 130\xb0+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0$

$\Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=180\xb0-130\xb0=50\xb0$ .....(3)

In âˆ†BOC,

∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)

$\Rightarrow \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{50\xb0}{2}+\angle \mathrm{BOC}=180\xb0\left[\mathrm{Using}\left(3\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BOC}=180\xb0-25\xb0=155\xb0\phantom{\rule{0ex}{0ex}}$

If the measure of one of the angles of a triangle is 130°, then the angle between the bisectors of the other two angles is _____155º_____.

#### Page No 11.28:

#### Question 5:

An exterior angle of a triangle is 105° and its two interior opposite angles are equal. The measure of each of these two angles is __________.

#### Answer:

Let the measure of each of the two interior opposite angles be *x*.

Measure of exterior angle of triangle = 105°

We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.

$\therefore x+x=105\xb0$

$\Rightarrow 2x=105\xb0$

$\Rightarrow x=\frac{105\xb0}{2}=52.5\xb0$

An exterior angle of a triangle is 105° and its two interior opposite angles are equal. The measure of each of these two angles is _____52.5º_____.

#### Page No 11.29:

#### Question 6:

The number of triangles that can be drawn with the measure of each angle less than 60°, is __________.

#### Answer:

Let the three angles of the triangle be *x*, *y *and *z*.

It is given that, *x* < 60º, * y* < 60º and *z* < 60º

Now, *x* + *y* + *z* < 60º + 60º + 60º

⇒ *x* + *y* + *z* < 180º

Or Sum of the angles of the triangle < 180º

This is not possible as the sum of the angles of a triangle is always equal to 180º.

Thus, no triangle can be drawn with the measure of each angle less than 60°.

The number of triangles that can be drawn with the measure of each angle less than 60°, is ______0______.

#### Page No 11.29:

#### Question 7:

A triangle cannot have two ___________ angles.

#### Answer:

An angle whose measure is more than 90º but less than 180º is called an obtuse angle.

If two angles of a triangle are obtuse, then the sum of these two obtuse angles would be more than 180º.

So, the sum of the three angles of the triangle in this case would be more than 180º. This is not possible as the sum of three angles of a triangle is always equal to 180º.

Thus, a triangle cannot have two obtuse angles.

A triangle cannot have two _____obtuse _____ angles.

#### Page No 11.29:

#### Question 8:

All angles of a triangle can be __________ angles.

#### Answer:

An angle whose measure is less than 90º is called an acute angle. A triangle can have three acute angles such that their sum is 180º. A triangle which has all acute angles is called an acute triangle.

All angles of a triangle can be ____acute____ angles.

#### Page No 11.29:

#### Question 9:

In a âˆ†*ABC*, if ∠*A* < ∠*B* < 45°, then âˆ†*ABC* is a/an* ________ *triangle.

#### Answer:

In âˆ†ABC, ∠A < ∠B < 45°. This means that the measure of ∠A is less than 45º and measure of ∠B is less than 45º. Also, the measure of ∠A is less than measure of ∠B.

So, ∠A < 45° and ∠B < 45°

∴ ∠A + ∠B < 45° + 45°

⇒ ∠A + ∠B < 90°

Adding ∠C to both sides, we have

∠A + ∠B + ∠C < 90° + ∠C .....(1)

We know

∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

So,

180° < 90° + ∠C [Using (1)]

Or 90° + ∠C > 180°

⇒ ∠C > 180° − 90°

⇒ ∠C > 90°

A triangle with one angle an obtuse angle is known as an obtuse triangle. So, âˆ†ABC is an obtuse triangle.

In a âˆ†ABC, if ∠A < ∠B < 45°, then âˆ†ABC is a/an ____obtuse____ triangle.

#### Page No 11.29:

#### Question 10:

In a triangle *ABC*, if ∠*A* > ∠*B* > ∠C and the measures of ∠*A*, ∠*B* and ∠*C* in degrees are integers, then the least possible values of *A, B* and *C *are _______ and ______ respectively.

#### Answer:

It is given that, in âˆ†ABC, ∠A > ∠B > ∠C and the measures of ∠A, ∠B and ∠C in degrees are integers.

So, the least value of ∠C is 1º and â€‹the least value of ∠B is 2º.

In âˆ†ABC,

∠A + ∠B + ∠C = 180° (Angle sum property of triangle)

∴ ∠A + 2° + 1° = 180°

⇒ ∠A = 180° − 3° = 177°

In a triangle ABC, if ∠A > ∠B > ∠C and the measures of ∠A, ∠B and ∠C in degrees are integers, then the least possible values of A, B and C are ____177°, 2°____ and ____1°___ respectively.**Note**: The value of ∠A depends upon the values of ∠B and ∠C. The least value of ∠A would be 61º. But, it that case the values of ∠B or ∠C would not be least.

#### Page No 11.29:

#### Question 11:

The measures of three angles of a triangle are in the ratio 1 : 2 : 3. Then, the triangle is a __________ triangle.

#### Answer:

Let the measure of three angles of the triangle be *x*, 2*x* and 3*x*.

Now,

$x+2x+3x=180\xb0$ (Angle sum property of triangle)

$\Rightarrow 6x=180\xb0$

$\Rightarrow x=\frac{180\xb0}{6}=30\xb0$

$\therefore 2x=2\times 30\xb0=60\xb0$ and $3x=3\times 30\xb0=90\xb0$

So, the measure of the angles of the triangle are 30º, 60º and 90º.

Now, a triangle with one angle a right angle is called a right triangle. Thus, the triangle is a right triangle.

The measures of three angles of a triangle are in the ratio 1 : 2 : 3. Then, the triangle is a _____right_____ triangle.

#### Page No 11.29:

#### Question 12:

The internal bisectors of ∠*B* and ∠*C* of âˆ†*ABC *meet at *O*. If *B + C *= 100°, then ∠*BOC *= __________.

#### Answer:

In âˆ†ABC, ∠B + ∠C = 100°.

Also, OB and OC are the bisectors of ∠B and ∠C, respectively.

$\therefore \angle \mathrm{OBC}=\frac{\angle \mathrm{B}}{2}.....\left(1\right)$

Similarly, $\angle \mathrm{OCB}=\frac{\angle \mathrm{C}}{2}.....\left(2\right)$

In âˆ†BOC,

∠OBC + ∠OCB + ∠BOC = 180º (Angle sum property of triangle)

$\Rightarrow \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}+\angle \mathrm{BOC}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{100\xb0}{2}+\angle \mathrm{BOC}=180\xb0\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BOC}=180\xb0-50\xb0=130\xb0\phantom{\rule{0ex}{0ex}}$

The internal bisectors of ∠B and ∠C of âˆ†ABC meet at O. If B + C = 100°, then ∠BOC = ______130º______.

#### Page No 11.29:

#### Question 1:

Define a triangle.

#### Answer:

A plane figure bounded by three lines in a plane is called a triangle. A triangle has three sides, three angles and three vertices. The figure below represents a ΔABC, with AB, BC and CA as the three sides; ∠A, ∠B and ∠C as the three angles; A, B and C as the three vertices.

#### Page No 11.29:

#### Question 2:

Write the sum of the angles of an obtuse triangle.

#### Answer:

In the given problem, Δ*ABC* is an obtuse triangle, withas the obtuse angle.

So, according to “the angle sum property of the triangle”, for any kind of triangle, the sum of its angles is 180°. So,

Therefore, sum of the angles of an obtuse triangle is.

#### Page No 11.29:

#### Question 3:

In Δ *ABC*, if ∠*B* = 60°, ∠*C* = 80° and the bisectors of angles ∠*ABC* and ∠*ACB* meet at a point *O*, then find the measure of ∠*BOC*.

#### Answer:

In Δ*ABC*,,and the bisectors of and meet at *O*.

We need to find the measure of

Since,*BO* is the bisector of

Similarly,*CO* is the bisector of

Now, applying angle sum property of the triangle, in Δ*BOC*, we get,

Therefore,.

#### Page No 11.29:

#### Question 4:

If the angles of a triangle are in the ratio 2 : 1 : 3, then find the measure of smallest angle.

#### Answer:

In the given problem, angles of Δ*ABC** *are in the ratio 2:1:3

We need to find the measure of the smallest angle.

Let,

According to the angle sum property of the triangle, in Δ*ABC*, we get,

Thus,

Since, the measure of is the smallest of all the three angles.

Therefore, the measure of the smallest angle is .

#### Page No 11.29:

#### Question 5:

State exterior angle theorem.

#### Answer:

Exterior angle theorem 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.

Thus, in Δ*ABC*

#### Page No 11.29:

#### Question 6:

The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.

#### Answer:

In the given problem, the sum of two angles of a triangle is equal to its third angle.

We need to find the measure of the third angle.

Thus, it is given, in

* * ........(1)

Now, according to the angle sum property of the triangle, we get,

(Using 1)

Therefore, the measure of the third angle is.

#### Page No 11.29:

#### Question 7:

In the given figure, if *AB* || *CD*, *EF* || *BC*, ∠*BAC* = 65° and ∠*DHF* = 35°, find ∠*AGH*.

#### Answer:

In the given figure,,,and

We need to find

Here, *GF *and *CD *are straight lines intersecting at point *H*, so using the property, “vertically opposite angles are equal”, we get,

Further, asand *AC *is the transversal

Using the property, “alternate interior angles are equal”

Further applying angle sum property of the triangle

In Δ*GHC*

Hence, applying the property, “angles forming a linear pair are supplementary”

As *AGC *is a straight line

Therefore,

#### Page No 11.29:

#### Question 8:

In the given figure, if *AB* || *DE* and *BD* || *FG* such that ∠*FGH* = 125° and ∠*B* = 55°, find *x* and* y*.

#### Answer:

In the given figure,,,and

We need to find the value of *x *and* y*

Here, asand *BD *is the transversal, so according to the property, “alternate interior angles are equal”, we get

Similarly, as and DF is the transversal

(Using 1)

Further, *EGH* is a straight line. So, using the property, angles forming a linear pair are supplementary

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

In with as its exterior angle

Thus,

#### Page No 11.30:

#### Question 9:

If the angles *A*, *B* and *C* of Δ*ABC* satisfy the relation *B* − *A* = *C* − *B*, then find the measure of ∠*B*.

#### Answer:

In the given Δ*ABC*,

,and satisfy the relation

We need to fine the measure of.

As,

........(1)

Now, using the angle sum property of the triangle, we get,

(Using 1)

Therefore,

#### Page No 11.30:

#### Question 10:

In Δ*ABC*, if bisectors of ∠*ABC* and ∠*ACB* intersect at *O* at angle of 120°, then find the measure of ∠*A*.

#### Answer:

In the given Δ*ABC*,, the bisectors of and meet at *O *and

We need to find the measure of

So here, using the corollary, “if the bisectors of and of a meet at a point *O*, then”

Thus, in Δ*ABC*

Thus,

#### Page No 11.30:

#### Question 11:

If the side *BC* of Δ*ABC* is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠*A*.

#### Answer:

In the given problem, we need to find the difference between the sum of the exterior angles and.

Now, according to the exterior angle theorem

.........(1)

Also,

.........(2)

Further, adding (1) and (2)

.........(3)

Also, according to the angle sum property of the triangle, we get,

.........(4)

Now, we need to find the difference between the sum of the exterior angles and.

Thus,

(Using 4)

Therefore,

#### Page No 11.30:

#### Question 12:

In a triangle *ABC*, if *AB* = *AC* and *AB* is produced to *D* such that *BD* = *BC*, find ∠*ACD*: ∠*ADC*.

#### Answer:

In the given ,and *AB *is produced to *D* such that

We need to find

Now, using the property, “angles opposite to equal sides are equal”

As

........(1)

Similarly,

As

........(2)

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

In Δ*BDC*

(Using 2)

From (1), we get

.......(3)

Now, we need to find

That is,

(Using 3)

(Using 2)

Eliminating from both the sides, we get 3:1

Thus, the ratio of is

#### Page No 11.30:

#### Question 13:

In the given figure, side *BC* of Δ*ABC* is produced to point D such that bisectors of ∠*ABC* and ∠*ACD** *meet at a point *E*. If ∠B*AC* = 68°, find ∠*BEC*.

#### Answer:

In the given figure, bisectors of and meet at *E* and

We need to find

Here, using the property: an exterior angle of the triangle is equal to the sum of the opposite interior angles.

In Δ*ABC *with as its exterior angle

........(1)

Similarly, in Δ*BE** *with as its exterior angle

(*CE *and *BE* are the bisectors of and)

........(2)

Now, multiplying both sides of (1) by

We get,

........(3)

From (2) and (3) we get,

Thus,

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