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Page No 236:

Question 1:

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

Answer:

A+B+C=180°      The sum of three angles of a triangle is 180°.55°+40°+C=180°95°+C=180°C=180°-95°C=85°
 

Page No 236:

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 xº, 2xº and 3xº. Then,

 x+2x+3x=180      The sum of three angles of a triangle is 180°6x=180x=30

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

Page No 236:

Question 3:

The angles of a triangle are (x − 40)°, (x − 20)° and 12x-10°. Find the value of x.

Answer:

Given angles are

 x-40+x-20+12x-10=18052x=180+7052x=250x=250×25x=100

Hence, the value of x is 100°.

Page No 236:

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)°.

 x+x+x+30=180     Sum of the three angles of a triangle is 180°3x+30=1803x=150x=50

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

Page No 236:

Question 5:

Prove that a triangle must have at least two acute angles.

Answer:

Let us consider different cases:

Case I: When two angles are 90°.
We know,
A+B+C=180°A+90°+90°=180°A=0°
Thus, no triangle will be formed.

Case II: When one angle is 90° and other is obtuse.
A=90° and B= ObtuseA+B+C=180°90°+B+C=180°C=90°-BC<0   B= Obtuse
Thus, the triangle will be not be formed.

Case III: When two angles are acute.
When two angles are acute, then the sum of two angles is less than 180, so that the third angle can be an acute or obtuse angle.

Page No 236:

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 236:

Question 7:

How many triangles can be drawn having its angles as 45°, 64° and 72°? Give reason for your answer.

Answer:

Given that, the angles of a triangle are 45°, 64°, and 72°.
We know that the sum of all the interior angles of a triangle should be 180°.

But, according to the question, we have the angles of 45°, 64°, and 72°.
Sum of the angles = 45° + 64° + 72° = 181° > 180° .
Thus, the angles do not satisfy the angle sum property of a triangle.

Hence, no triangle can be drawn having angles of 45°, 64°, and 72°.

Page No 236:

Question 8:

How many triangles can be drawn having its angles as 53°, 64° and 63°? Give reason for your answer.

Answer:

Ans

Page No 236:

Question 9:

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°]         

 x+x+10+x+20=180     Sum of the three angles of a triangle is 180°3x+30=1803x=150x=50

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

 

Page No 236:

Question 10:

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 B and C

A+B+C=180°72°+2OBC+2OCB=180°       Sum of the three angles of a triangle is 180°2OBC+OCB=108°OBC+OCB=54°180°-BOC=54°        Since, OBC+OCB+BOC=180°BOC=126°

Hence magnitude of BOC is 126°.

Page No 236:

Question 11:

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

BOC=90°+12A

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 236:

Question 12:

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

BOC=90°+12A

 135°=90°+12A45°=12AA=90°

Hence the triangle is a right angled triangle.

Page No 236:

Question 13:

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.

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

BOC=90°+12A

 120°=90°+12A30°=12AA=60°

are equal as it is given that .

A+B+C=180°     Sum of three angles of a triangle is 180°60°+2B=180°          ABC=ACBB=60°

Hence, .

Page No 236:

Question 14:

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 237:

Question 15:

In the given figure, if ∆XYZ and ∆LMN are equilateral triangles, then find the measure of ∠ZAB.

Answer:

We have,
XYZ=60°             Each angle of an equilateral triangle is 60°

Now, XYP+XYZ+ZYL=180°      Linear Pair65°+60°+ZYL=180°  ZYL=55°Also,  MLN=60°  MLQ+MLN+NLY=180°      Linear Pair75°+60°+NLY=180°NLY=45°In BYL55°+45°+YBL=180°  YBL=80°YBL=ABZ=80°           Vertically opposite anglesIn ABZ, we haveZAB+ABZ+AZB=180°  ZAB+80°+60°=180°  XYZ is equilateralZAB=40°

Thus, the measure of  ∠ZAB is 40°.



Page No 244:

Question 1:

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 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.

Page No 244:

Question 2:

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 244:

Question 3:

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 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,.

Page No 244:

Question 4:

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 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 No 244:

Question 5:

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 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,



Page No 245:

Question 6:

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 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,.

Page No 245:

Question 7:

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 ΔDCE

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

Therefore,.

Page No 245:

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 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

BQC=90°-12A        ......2                

Adding (1) and (2), we get

Thus,

Hence proved.

Page No 245:

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 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 No 245:

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,

BDA=BAD                .....1

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

DAB=y, BAC=3y

Thus, 

y+3y=DAC4y=72°y=72°4y=18°

Hence, DAB=18°, BAC=3×18°=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,

EAC=ADC+x108°=18°+xx=90°

Therefore,.

Page No 245:

Question 11:

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 246:

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 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,.

Page No 246:

Question 13:

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

Answer:

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

Page No 246:

Question 14:

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

Answer:

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.

Page No 246:

Question 15:

In ΔABC, 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:

Given that, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q
To prove: ∠BPC + ∠BQC = 180°.

We have,
ABC+CBD=180°12ABC+CBD=90°PBC+QBC=90°  PB and QB are the bisectors of ABC and CBDPBQ=90°    .....1Similarly, PCQ=90°   .....2Since, the sum of angles of quadrilateral PBCQ is 360°.PBQ+BQC+PCQ+BPC=360°BPC+BQC=180°    From 1 and 2

Hence proved.

Page No 246:

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 246:

Question 2:

Write the sum of the angles of an obtuse triangle.

Answer:

In the given problem, ΔABC 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 247:

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 ΔABC,,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 ΔBOC, we get,

Therefore,.

Page No 247:

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 ΔABC 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 ΔABC, 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 247:

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 ΔABC

Page No 247:

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 247:

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 ΔGHC

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

As AGC is a straight line

Therefore,

Page No 247:

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 247:

Question 9:

If the angles A, B and C of ΔABC satisfy the relation BA = CB, then find the measure of ∠B.

Answer:

In the given ΔABC

,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 247:

Question 10:

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

Answer:

In the given ΔABC,, 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 ΔABC

Thus,

Page No 247:

Question 11:

If the side BC of ΔABC 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 247:

Question 12:

In a triangle ABC, if ABAC and AB is produced to D such that BDBC, 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 ΔBDC

(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 247:

Question 13:

In the given figure, side BC of ΔABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 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 ΔABC with as its exterior angle

    ........(1)

Similarly, in ΔBE 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,

Page No 247:

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 5x, 3x and 7x.

Now,

5x+3x+7x=180°            (Angle sum property of triangle)

15x=180°

x=180°15=12°

5x=5×12°=60°, 3x=3×12°=36° and 7x=7×12°=84°

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 247:

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 2x, 4x and 3x.

Now,

2x+4x+3x=180°            (Angle sum property of triangle)

9x=180°

x=180°9=20°

2x=2×20°=40°, 4x=4×20°=80° and 3x=3×20°=60°

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 248:

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 248:

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 ∆ABC, ∠A = 130º.

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

OBC=B2      .....1

Similarly, OCB=C2              .....2

Now,

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

130°+B+C=180°

B+C=180°-130°=50°         .....(3)

In ∆BOC,

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

B2+C2+BOC=180°              Using 1 and 2B+C2+BOC=180°50°2+BOC=180°                        Using 3BOC=180°-25°=155°

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 248:

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.

x+x=105°

2x=105°

x=105°2=52.5°

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 ___52.5º___.

Page No 248:

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 xy and z.

It is given that, 

x < 60º,  y < 60º and z < 60º

Now, 

xyz < 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 248:

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 248:

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 248:

Question 9:

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

Answer:


In ∆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, ∆ABC is an obtuse triangle.

In a ∆ABC, if ∠A < ∠B < 45°, then ∆ABC is a/an __obtuse__ triangle.
 

Page No 248:

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 ∆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 ​the least value of ∠B is 2º.

In ∆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 248:

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, 2x and 3x.

Now,

x+2x+3x=180°            (Angle sum property of triangle)

6x=180°

x=180°6=30°

2x=2×30°=60° and 3x=3×30°=90°

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 248:

Question 12:

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

Answer:



In ∆ABC, ∠B + ∠C = 100°.

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

OBC=B2      .....1

Similarly, OCB=C2              .....2

In ∆BOC,

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

B2+C2+BOC=180°              Using 1 and 2B+C2+BOC=180°100°2+BOC=180°                        GivenBOC=180°-50°=130°


The internal bisectors of ∠B and ∠C of ∆ABC meet at O. If B + C = 100°, then ∠BOC = ____130º____.



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