Page 8.43

### Exercise 8.3

Type 1

#### Question 1

**Fill up the blanks:**

**(i) In a triangle sum of measures of three angles is __**

**(ii) In a triangle , maximum number of acute angles is __**

**(iii) In a triangle, minimum number of acute angles is __**

**(iv) In figure (i) , if ∠ACD=130° , ∠ABC=48° , then ∠BAC=__**

**(v) If a side of a triangle is extended , then the exterior angle so formed is equal to the __ of the opposite interior angles .**

#### Question 2

**Which of the following statements are true (T) and which are false (F):**

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

**(ii) Sum of the four angles of a quadrilateral is three right angles.**

**(iii) In a triangle, there may be two right angles.**

**(iv) In a triangle, exterior angle is smaller than each interior opposite angle.**

**(v) In a triangle, there may be two obtuse angles.**

**(vi) In a triangle, there may be two acute angles.**

**(vii) In a triangle, exterior angle is equal to the sum of interior opposite angles.**

#### Question 3

**(i) In a triangle, one angle measures 70°,then write the sum of two remaining angles in degrees**

**(ii) In a ΔABC , ∠C=40° ,∠B=80° , find the value of ∠A **

**(iii) In a ΔABC , ∠B=105° and ∠C=50° , find the value of ∠A **

#### Question 4

**Give reasons for your answer in each of the following cases. Can in a triangle there be ?**

**(i) Two right angles**

**(ii) Two obtuse angles**

**(iii) Two acute angles**

**(iv) Each angle greater than 60°**

**(v) Each angle smaller than 60°**

**(vi) Each angle equal to 60°**

#### Question 5

**(i) At least how many acute angles are there in a triangle ?**

**(ii) What Will be the maximum numbers of acute angles in a triangles ?**

**(iii) At most how many obtuse angles can there be in a triangle ?**

#### Question 6

**(i) If α be an angle of a triangle such that 90°<α<180° , then what type of triangle will it be ?**

**(ii) In a right angled triangle, how many acute angles are there ?**

#### Question 7

**(i) In a right angled triangle, one acute angle is 40° , then find the measure of the other acute angle in degrees.**

**(ii) A quadrilateral has three angles as 110° , 40° and 50° . Find the value of the fourth angle.**

#### Question 8

**How many right angles is the sum of the interior angles of a rhombus ?**

Sol :

#### Question 9

**(i) In the given figure side BC is produced to M, ∠ACM=100° and ∠ABC=45° , find ∠BAC.**

Sol :

**(ii) In the given figure, if ∠MBC=140° , ∠BCA=40° , find ∠BAC.**

**(iii) If θ be the exterior angle of a triangle and sum of the two opposite interior angles, one is β , then what is the relation between θ and β ?**

**(iv) If in ΔABC (figure below) , ∠ACD=105° and ∠BAC=35° , find the value of ∠ABC.**

<fig to be added>

#### Question 10

(i) In the following figures, find the values of each angle.

(a)

(b)

**(ii) In the given figure, AM and DM are the bisectors of ∠A and ∠D respectively of quadrilateral ABCD . Find the value of ∠AMD in degrees.**

Page 8.45

#### Question 11

**(i) In a triangle, two angles are equal and third angle exceeds each of these angles by 30°. Determine all angles of the triangle**

**(ii) In a triangle, an exterior angle is 115° and one of the opposite interior angles 35° , then find the remaining angles **

**(iii) An angle of a triangle measures 65° , then find the other two angles if their difference is 25°**

**(iv) In a right angled triangle, the bigger acute angle is two times the smaller acute angle, then find the measure of bigger acute angle.**

**(v) In a triangle, one of the three angles twice the smallest angle and other is three times the smallest angle. Find the angles.**

**(vi) Sum of two angles of a triangle is 80° and their difference is 20° . Find all angles of the triangle .**

#### Question 12

**Find the angle formed by the bisectors of two acute of a right-angled triangle.**

Sol :

#### Question 13

**(i) If the angles of a triangle are in the ratio 2:3:4 , then find the values of the biggest and smallest angles.**

**(ii) If the angles of a triangle are in the ratio 1:2:3 , then find the value of the greater angle of the triangles.**

**(iii) If angles of a triangle are in the ratio 2:3:5 , then find the measure in degree of the smallest angle.**

**(iv) if angles of a quadrilateral are in the ratio 1:2:3:4 , then find the angles of the quadrilateral**

Type 2

#### Question 14

**(i) In figure (i) below , side QR of a ΔPQR is produced to S . If ∠P:∠Q:∠R=3:2:1 and RT⊥PR , find ∠TRS**

Sol :

**(ii) In quadrilateral ABCD, AD||BC and bisectors of interior ∠A and ∠B meet at O , find the measure of ∠AOB in degree [fig. (ii)]**

Sol :

**(iii) In figure (iii) , AB||CD , then find the value of α+β+γ.**

[Hint : Draw a line OE through point O parallel to AB or CD , then ∠BOE=180°-α [alternate angles]

∠DOE=180°-β [alternate angle]

Now , ∠BOE+∠DOE=γ=180°-α+180°-β

or α+β+γ=360°]

**(iv) In figure (iv) , l||m , find the value of x**

<fig to be added>

[Hint: l||m and they are intersected by a transversal

∴ 60°+y=180°⇒y=180°-60°=120°

∴ x=y+50°=120°+50°=170°]

**(v) In figure (v) , ABC is an isosceles triangles whose side AB=AC and XY||BC. If ∠A=30° , then find the value if ∠BXY in degrees.**

[Hint: AB=AC [∴ ΔABC is an isoceles triangle]

∴ ∠BXY+∠ABC=180° or ∠BXY + 75° = 180°

∴ ∠BXY=180°-75° = 105° ]

**(vi) In figure (a) and (b) , l||m , then find the value of x **

(a)

(b)

**(vii) In figure (a) and (b) below, find x and y as required.**

(a)

(b)

[Hint: In figure (a)

From ΔAEC, x=30°+42°=72°

Also, y=180°-42°=138°

In figure (b) exterior ∠CBD=30°+34°=64°=∠EBD

∴ Exterior angle x=∠EBD+∠EDB=64°+45°=109° ]

(viii) In the given figure, sides BC, CA and BA of ΔABC are produced to D, Q and P respectively. If ∠ACD=100° and ∠QAP=35° , then find all angles of the triangle.

[Hint: ∠BAC=∠QAP=35°

Now , exterior ∠DCA=∠CAB+∠ABC

⇒100°=35°+∠ABC or ∠ABC=100°-35°=65°

Also , ∠ACB=180°-100°=80° ]

(ix) In ΔABC , ∠B=45° , ∠C=55° and bisector of ∠A meets BC in D. Find ∠ADB and ∠ADC

[Hint: ∠A=180°-(45°+55°)=80°

∵ AD is the bisector of ∠A ,

∴ ∠BAD

Now exterior ∠ADB=∠DAC+∠DCA=40°+55°=95°

and exterior ∠ADC=∠ABD+∠BAD=45°+40°=85° ]

Type 3

#### Question 15

**In the given figure , prove that P||m**

#### Question 16

**Prove that the sum of the exterior angles formed on producing sides of a triangle in the same order is 360° or four right angles .**

#### Question 17

**Prove that each angle of a triangle will be 60°, if all its angles are of the same measure.**

Sol :

#### Question 18

**If an angle of a triangle is equal to the sum of remaining two angles, prove that triangle is right angled triangle.**

Sol :

#### Question 19

**In ΔABC,∠A = 40°, if bisectors of ∠B and ∠C meet at point O, then prove ∠BOC=110°**

Sol :

[Hint: See the given figure

Since BO and CO are bisectors of ∠ABC

and ∠ACB respectively ,

∴ ∠ABO=∠OBC=∠1 (say)

and ∠ACO=∠OCB=∠2 (say)

Now , in ΔOBC,

∠1+∠2+∠BOC=180°

In ΔABC , ∠ABC+∠ACB+∠BAC=180°..(i)

or 2∠1+2∠2+40°=180°..(ii) [∵ ∠BAC=40°]

∴ ∠1+∠2

∴ From (1) , 70°+∠BOC=180°

∠BOC=180°-70°=110°]

#### Question 20

**In right angled ΔABC, ∠A=90° , and bisectors of ∠B and ∠C meet at O, prove that ∠BOC = 135°**

[Hint: Proceed as in question 19

∠1+∠2+∠BOC=180°

Also , ∠A+∠B+∠C=180°

90°+2∠1+2∠2=180°

⇒∠1+∠2

From (i) , ∠BOC=180°-(∠1+∠2)

=180°-45°

=135°]

#### Question 21

**In a triangle ABC, ∠A=90° and AL⊥BC , prove that ∠BAL=∠ACB **

<fig to be added>

[Hint : Draw figure yourself, in ΔABC , ∠ABC ,∠CAB+∠B+∠ACB=180°..(i)

Now, in ΔALB , ∠ALB=90°

∴ ∠ALB+∠B+∠BAL=180°..(ii)

From (i) and (ii) , we get

∠ALB+∠B+∠BAL=∠CAB+∠B+∠ACB

⇒90°+∠B+∠BAL=90°+∠B+∠ACB

⇒∠BAL=∠ACB ]

#### Question 22

**If a transversal intersects two parallel lines, prove that bisectors of two interior angles on the same side form 90° with one another.**

<fig to be added>

[Hint: AB||CD and they are intersected by transversal EF at M and N (see the given fig)

Since sum of the interior angles on the same side of transversal is 180°

∴ ∠BMN+∠DNM=180°

⇒

Now MO and NO are bisectors of ∠BMN and ∠DNM respectively.

∴ ∠OMN+∠ONM=90° or ∠1-∠2=90°

Now , ∠1+∠2+∠MON=180°

⇒90°+∠MON=180°

⇒∠MON=90°]

#### Question 23

**Prove that the sum of interior angles of a hexagon is 720°**

[Hint: We know that the sum of the angles subtended by sides at the centre of a polygon is 360°.

Angles subtended a side of a regular polygon at its centre where n is the number of sides which is here 6

∵ ∠AOB = 60°; Also AO=BO=AB

∴ ΔAOB is an equilateral triangle.

∴ ΔAOB=∠OBA=∠OAB=60°

Similarly, ΔOAF is also an equilateral triangle.

∴ ∠OAB+∠OAF=60°+60°=120°=∠FAB=an interior angle

∴ Sum of all the interior angles of a hexagon =6×120°=720]

#### Question 24

**In a triangle ABC , CD is perpendicular to side AB and BE is perpendicular to side AC. Prove that ∠ABE=∠ACD**

[Hint: In ΔABE and ΔADC

∠AEB=∠ADC; ∠A=∠A

∴ ∠ABE=∠ACD]

#### Question 25

**In ΔABC , internal bisectors of ∠B and ∠C meet at P and external bisectors of these angles meet at Q. Prove that**

**∠BPC+∠BQC=180°**

[Hint: ∠BPC=90°+ and

∠BQC=90°-

∴ ∠BPC+∠BQC

=90°++90°-

=180°]

#### Question 26

**In the given figure (i) , two plane mirrors m and n are perpendicular to each other. Show that incident ray CA on being reflected is parallel to reflected ray BD.**

[Hint: AP and BP are normals to mirrors *m* and *n* respectively ]

Angle of incidence = Angle of reflection [Law of reflection]

Now in figure, (ii) ∠CAP=∠PAB=x

∠ABP=∠PBD=y

In ΔAPB , ∠APB=90°

∴ ∠PAB+∠PBA=90°

∴ x+y=90° or 2(x+y)=2×90°

or 2x+2y=180° or ∠CAB+∠ABD=180°

Since these angles are on the same side of the transversal AB, therefore CA||BD

#### Question 27

**In the given figure, PS is bisector of ∠P and PT⊥QR , prove that **

[Hint: Let ∠QPS=∠RPS=x and ∠TPS=y

In right angled ΔPQT , ∠QPT+∠Q=(x-y)+∠Q=90°..(i)

In right angled ΔPRT , ∠RPT+∠R=90° or

(x+y)+∠R=90°..(iii) [∵ ∠RPT=∠RPS+∠TPS=x+y]

From (i) and (ii) , x-y+∠Q=x+y+∠R

⇒ ∠Q-∠R=2y

∴ y=∠TPS]

#### Question 28

**In the given figure, side BC of ΔABC is produced so as to form the ray BD and CE||AB. Without using the theorem related to sum of angles of a triangle, prove that:**

**∠ACD=∠A+∠B and hence determine ∠A+∠B+∠C=180°.**

<fig to be added>

[Hint: ∵ AB||CE and transversal AC meets them

∴ ∠1=∠4 [alternate angles]

and ∠2=∠5 [corresponding angles]

∴ ∠1+∠2=∠4+∠5=∠ACE+∠ECD=∠ACD

∴ ∠A+∠B=∠ACD

Again , ∠A+∠B+∠C=∠ACD+∠C=180° [∵ Line AC stands on ray BCD ∠C+∠ACD=180°]

#### Question 29

**In ΔABC , side BC is produced to D and bisector of ∠A meets BC at L. Prove that ∠ABC+∠ACD=2∠ALC**

<fig to be added>

[Hint: ∠BAL=∠CAL=∠1 (say) [∵ AL is bisector of ∠BAC]

In ΔABL , ∠ALC=∠B+∠1

or 2∠ALC=2∠B+2∠1..(i)

In ΔABC, exterior ∠ACD=∠B+∠A=∠B+2∠1

∴ ∠B+∠ACD=∠B+∠B+2∠1=2∠B+2∠1..(ii)

From (i) and (iii) , we get

2∠ALC=∠B+∠ACD=∠ABC+∠ACD ]

#### Question 30

**ABCDE is a regular pentagon and bisector of ∠A meets the side CD at point . Prove that ∠AMC=90°**

<fig to be added>

[Hint: We know that a regular pentagon has all its five sides equal. Also angle subtended by any side at the centre

∴ ∠COD=72°

In ΔOCD, OC=OD

∴ ∠OCD=∠ODC

∴ ∠OCF=180°-54°=126°

Consider ΔAOB and ΔAOE

AB=AE; ∠BAO=∠EAO [∵ AO is bisector of ∠BAE]

AO=AO

∴ ΔAOB≅ΔAOE; BO=OE

Now , AO will bisect ∠COD

∴ ∠COM=∠MOD

∴ ∠OMC=∠OCM+∠COM=54°+36°=90°

Hence, ∠AMC=90°]

#### Question 31

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

<fig to be added>

[Hint: AE is bisector of ∠CAD

∴ ∠CAE=∠DAE=∠1 (say)

Now exterior ∠CAD=∠B+∠C⇒2∠1

But ∠B+∠C=2∠C [∵ ∠B=∠C]

∴ 2∠C=2∠1 or ∠C=∠1=∠CAE

∵ These are alternate angles

∴ AE||BC ]

#### Question 32

**If arms of an angle are perpendicular to the arms of another angle, then prove that the angles will either be equal or supplementary**

[Hint: In figure (i) , ∠BFO=∠EDO=90°

∠BFO=∠EOD [Vertically opposite angles]

∴ ∠BFO+∠BOF=∠EDO+∠EOD

or 180°-∠FBO=180°-∠OED

or ∠FBO=∠OED

i.e., ∠ABC=∠FED

In figure (ii) , ∠B+∠D+∠E+∠F=360°

or ∠B+90°+∠E+90°=360°

or ∠B+∠E=180° ]

#### Question 33

**In the given figure, bisectors of ∠B and ∠D meet produced CD and AB at P and Q respectively.**

**Prove that ∠P+∠Q**

[Hint: Let ∠ABC=2x and ∠ADC=2y, then in quadrilateral PBQD, we have

∠P+∠PDQ+∠Q+∠QBP=360°

or ∠P+(180°-y)+∠Q+(180°-x)=360°

or ∠P+∠Q=x+y

[(∵ 2x=∠ABC ∴ and 2y=∠ADC

∴ )]