RD Sharma 2018 Solutions for Class 9 Math 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 Math Triangle And Its Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2018 Book of class 9 Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2018 Solutions. All RD Sharma 2018 Solutions for class 9 Math 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:

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 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 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 11.10:

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 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)°.

 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 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

A=B+C      Since, one angle is sum of the other two A+B+C=180°   Sum of the three angles of a triangle is 180°A+A=180°2A=180°A=90°

 

Hence, the given triangle is a right angled triangle.

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°]         

 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 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 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 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

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 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

BOC=90°+12A

 135°=90°+12A45°=12AA=90°

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.

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 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 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 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 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 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 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 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 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 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 ΔDCE

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 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 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 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 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 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 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,

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 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 = 12A.

Answer:

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

We need to prove:

Now, using the exterior angle theorem,

ABE=BAC+ACB   .….(1)

 As ABE and ACB are bisected

DCB=12ACB

Also,

DBA=12ABE

Further, applying angle sum property of the triangle

In  ΔDCB

CDB+DCB+CBD=180°CDB+12ACB+DBA+ABC=180°

  CDB+12ACB+12ABE+ABC=180°      .....2

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

ABC+ABE=180°ABC+12ABE+12ABE=180°ABC+12ABE=180°-12ABE            .....3

So, using (3) in (2)

CDB+12ACB+180°-12ABE=180°CDB+12ACB-12ABE=0CDB=12ABE-ACBCDB=12CABD=12A

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 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 11.22:

Question 13:

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 11.22:

Question 14:

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 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:

Define a triangle.

Answer:

A plane figure bounded by three lines in a plane is called a triangle. The figure below represents a ΔABC, with AB, AC and BC as the three line segments.

Page No 11.23:

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 11.23:

Question 3:

In Δ ABC, if u∠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 11.23:

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 11.23:

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 11.23:

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 11.23:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.24:

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 11.25:

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 ΔABC we are given that the three angles are equal. So,

According to the angle sum property of a triangle, in ΔABC

 

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

So, the correct option is (c).

Page No 11.25:

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 ΔABC

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

()

Therefore, the correct option is (b).

Page No 11.25:

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 ΔABC, CD is the ray extended from the vertex C of ΔABC. 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 ΔABC

Therefore, each of the two opposite interior angles is

So, the correct option is (d).

Page No 11.25:

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 ΔABC

                  .........(2)

Further, using (2) in (1),

Thus,

Therefore, the correct option is (d).

Page No 11.25:

Question 5:

Side BC of a triangle ABC has been produced to a point D such that ∠ACD = 120°. If ∠B = 12 ∠A is equal to

(a) 80°

(b) 75°

(c) 60°

(d) 90°

Answer:

In the given problem, side BC of ΔABC 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 ΔABC

Also, (Using 1)

Thus,

Therefore, the correct option is (a).

Page No 11.25:

Question 6:

In ΔABC, ∠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 ΔABC, . 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 11.25:

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 ΔABC, 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 11.25:

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 ΔABC, 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 11.25:

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 ΔABC,, AD bisects and .

Here, we need to find.

As, AD bisects,

We get, 

Now, according to angle sum property of the triangle

In ΔABD

Hence,

Therefore, the correct option is (c).

Page No 11.25:

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 ΔABC, 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 ΔABC

Thus, 

Therefore, the correct option is (a).

Page No 11.25:

Question 11:

In a ΔABC, 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 ΔABC,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 ΔABC, we get,

Similarly, in ΔBOC

Hence,

Therefore, the correct option is (b).

Page No 11.25:

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 ΔODB, we get,

(using 1)

Thus,

Therefore, the correct option is (b).



Page No 11.26:

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 ΔOBD, we get, 

Thus,

Therefore, the correct option is (b).

Page No 11.26:

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 ΔAOC, we get,

Similarly, in ΔBOD

Thus,

Therefore, the correct option is (b).

Page No 11.26:

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 ΔABC 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 ΔABC, 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 11.26:

Question 16:

In the given figure, if ABBC. 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 ΔBOC 

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

In ΔEOC

Further solving for x, we get,

Thus,

Therefore, the correct option is (b).



Page No 11.27:

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 ΔABC, 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 11.27:

Question 18:

In the given figure, for which value of x is l1 || l2?

(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 ΔABC

Thus,

Therefore, the correct option is (d).

Page No 11.27:

Question 19:

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

(a) 32x

(b) 43x

(c) x

(d) 34x
 

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 ΔABC

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

Similarly, in ΔOCD

(using 1)

Thus,

Therefore, the correct option is (a).

Page No 11.27:

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 ΔABC

Thus,

Therefore, the correct option is (d).



Page No 11.28:

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 ΔABD

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 11.28:

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 ΔABC

Thus,

Therefore, the correct option is (c).

Page No 11.28:

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 ΔPRQ

Thus,

Therefore, the correct option is (a).

Page No 11.28:

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 ΔABC 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 ΔABC

Thus,

Therefore, the correct option is (c).



Page No 11.29:

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 ΔABC

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

Also, applying angle sum property of a triangle

In ΔAOC

Thus,

Therefore, the correct option is (c).

Page No 11.29:

Question 26:

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

(a) 90°+x°2

(b) 90°-x°2

(c) 180°+x°2

(d) 180°-x°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 ΔABC are produced to P and Q respectively and the bisectors of and intersect at O, therefore, we get,

Hence, in ΔABC

Thus,

Therefore, the correct option is (b).

Page No 11.29:

Question 27:

In a ΔABC, ∠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 ΔABC 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 11.29:

Question 28:

The side BC of ΔABC 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) 7212°

(c) 145°

(d) none of these

Answer:

In the given problem, BC of ΔABC 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 ΔABC

Now, as AL is the bisector of

Also, is the exterior angle of ΔALC

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 11.29:

Question 29:

In the given figure, if l1 || l2, the value of x is

(a) 2212

(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 ΔABC

(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 11.29:

Question 30:

In ΔRST (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 ΔABC meet at a point O, then

In ΔRST

Further solving for x, we get,

Thus,

Therefore, the correct option is (d).



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