S.chand books class 8 maths solution chapter 17 Triangles exercise 17 A

 EXERCISE 17 A

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Q1 | Ex-17A | Class 8| S.Chand | Composite maths | Triangles | myhelper

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Question 1

The three angles of a triangle measures (2x-10°) , (x+31°) and (5x+7°) . Find the value of x and hence all the angles of the triangle

Sol :

(2x-10°)+(x+31°)+(5x+7°)=180° (Angle sum property of triangle)

2x-10°+x+31°+5x+7°=180°

(2x+x+5x)+(-10°+31°+7°)=180°

8x+28°=180°

8x=180°-28°

$x=\frac{152}{8}=19$

x=19°

(2x-10°)=2(19°)-10°=38°-10°=28°

(x+31°)=(19°)+31°=50°

(5x+7°)=5(19°)+7°=95°+7°=102°

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Question 2

One of the exterior angles of a triangle is 153° and the interior opposite angles are in the ratio 5 : 4 . Find the three interior angles of the triangle.

Sol :

5x+4x=153° (An exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles)

9x=153°

$x=\frac{153}{9}$

x=17°

∠A=5x=5(17°)=85°

∠B=4x=4(17°)=68°

In ΔABC,

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

85°+68°+∠C=180°

153°+∠C=180°

∠C=180°-153°=27°

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Question 3

If the three angles of a triangle are (x+15°) , ($\dfrac{6x}{5}$+6°) and ($\dfrac{2x}{3}$+30°) , prove that the triangles is an equilateral triangles.

Sol :

(x+15°)+($\dfrac{6x}{5}$+6°)+($\dfrac{2x}{3}$+30°)=180° (Angle sum property of triangle)

$x+\frac{6x}{5}+\frac{2x}{3}+15+6+30=180$

$\frac{15x+3\times 6x+2x\times 5}{15}+51^{\circ}=180^{\circ}$

$\frac{15x+3\times 6x+2x\times 5}{15}$=180°-51°

$\frac{15x+3\times 6x+2x\times 5}{15}$=129°

$\frac{43x}{15}=129^{\circ}$

$x=\frac{129\times 15}{43}$

$x=\frac{1935}{43}$

x=45°

First angle=x+15=45+15=60°

Second angle$=\frac{6x}{5}+6=\frac{6\times 45}{5}+6$=54+6=60°

Third angle$=\frac{2x}{3}+30=\frac{2\times 45}{3}+30$=30+30=60°

All angles are equal, so its a equilateral triangle.

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Question 4

Find the value of x and y in the following f‌igures:

(i)

Sol :

∠ACD+∠ACB=180°

105°+∠ACB=180°

∠ACB=180°-105°=75°

In ΔABC,

x=70°+∠ACB (exterior angle is equal to sum of two opposite interior angles)

x=70°+75°=145°

(ii)

Sol :

In ΔABD,

∠ADC+∠ADE=180° (linear pair)

∠ADC+112°=180°

∠ADC=180°-112°=68°

∠ACD+∠ADC+∠CAD=180° (Angle sum property of triangle)

y+68°+52°=180°

y=180°-120°=60°

In ΔABD,

∠BAD+∠ABD+∠ADB=180° (angle sum property of triangle)

x+52°+48°+68°=180°

x+168°=180°

x=180°-168°=12°

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Question 5

In the given figure, ∠A = 54° , BO and CO are the bisectors of ∠B and angle C . Find ∠BOC [Hint: in ΔABC , 2x+2y+54°=180°➝2(x+y)=126°➝x+y=63°

In ΔBOC , x+y+z=180°]

Sol :

In ΔABC,

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

54°+(x+x)+(y+y)=180°

2x+2y=180°-54°

2(x+y)=126°

x+y=63°...(i)

In ΔBOC,

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

z+x+y=180°

z+63°=180° [x+y=63°]

z=180°-63°=117°

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Question 6

In an isosceles triangle each base angle is 30° greater than the vertical angle . Find the measure of all the three angles of the triangles .

Sol :

ATQ, Triangle is isosceles and each base angle is 30° greater than vertical angle.

Vertical angle be x

Base angle=(x+30°)

x+(x+30°)+(x+30°)=180° (Angle sum property of triangle)

x+x+30°+x+30°=180° (Angle sum property of triangle)

3x=180°-60°

x=120°

$x=\frac{120}{3}$

x=40°

Angles are x=40° 

x+30°=40°+30°=70°

x+30°=40°+30°=70°

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Question 7

In the figure given below AN=AC , ∠BAC=52° , ∠ACK=84°, and BCK is a straight line. Prove that NB = NC

Sol :

In ΔABC,

∠ACK=∠BAC+∠ABC (exterior angle is equal to sum of interior angle)

84°=52°+∠ABC 

∠ABC=84°-52°=32°........(i)

In ΔANC, AN=NC. So, its a isosceles triangle

∠NAC=∠NCA=52°

In ΔACN, (isosceles triangle)

∠NAC+∠ANC+∠NCA=180°

52°+x+x=180° (Angle sum property of triangle)

2x=180°-52°

2x=128°

$x=\frac{128}{2}$

x=64°

∠NCB=180°-(∠NCA-∠ACK)

∠NCB=180°-(64°+84°) (linear pair)

=180°-148°=32°....(ii)

From (i) and (ii)

∠ABC=32°=∠NCB

So, ΔNCB is isosceles then NB=NC

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Question 8

In the figure AB=AC . Prove that BD=BC

Sol :

Given AB=AC, 

To prove=BD=BC

AB=AC which means ΔABC is isosceles Δ, ∠ABC=∠ACB

In ΔABC

∠BAC+∠ABC+∠ACB=180° (Angle sum property of triangle)

40°+(30°+x)+(30°+x)=180°

40°+30°+x+30°+x=180°

100°+2x=180°

2x=180°-100°

2x=80°

x=40°

Also, ∠ABE=∠ACB=x+30°=40°+30°=70°

In ΔDBC,

∠DBC+∠BCD+∠CDB=180° (Angle property of triangle)

40°+70°+∠CDB=180°

∠CDB=180°110°=70°....(ii)

∠CDB=70°=∠BCD

So ,ΔDBC is isosceles then BD=BC

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Question 9

The ratio between the vertical angle and base angle of an isosceles triangle is 2 : 5. Find the angles of the triangle.
[Hint: $\dfrac{vertical~angle}{base~angle}=\dfrac{2}{5}$, i.e., $\dfrac{v}{b}=\dfrac{2}{5}\rightarrow v=\dfrac{2}{5}b$ (v denotes vertical angle and b denotes base angle)
Now , v+b+b=180°➝$\dfrac{2}{5}$b+b+b=180°]

Sol :

$\frac{\text{Vertical Angle}}{\text{Base Angle}}=\frac{2x}{5x}$ of a isosceles triangle

2x+5x+5x=180° (Angle sum property of triangle)

12x=180°

$x=\frac{180}{12}$

x=15°

Angles are 

2x=2×15=30°

5x=5×15=75°

5x=5×15=75°

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Question 10

Prove that the sum of the exterior angles of a triangle taken in order is 360°. i.e., x + y + z = 360°.

[Hint: Let int. ∠A = a° , int ∠B = b°, int.∠C=c°➝a + b + c = 180°

By ext. ∠Property x=a+b , y=b+c , z=a+c

x+y+z=a+b+b+c+a+c=2(a+b+c].

Sol :

∠BAC=180°-y

∠ABC=180°-z

∠ACB=180°-x

In ΔABC,

∠BAC+∠ABC+∠ACB=180° (Angle sum property of triangle)

180°-y+180°-z+180°-x=180°

-x-y-z+540°=180°

-(x+y+z)=180°-540°

-(x+y+z)=-360°

x+y+z=360°

Sum of exterior angle of triangle is 360°

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Question 11

Find the lettered angles in each of the following figures-

Sol :

(i)

In ΔABC, (isosceles because AB=AC)

So, ∠ABC=∠ACB=65°

Also, AD||BC (given) and AC is transversal

∠BCA=∠DAC=65° (alternate interior angle)

∠DAC=∠DCA

65°=x

∵ In ΔADC, (isosceles because AD=DC)

∠DAC+∠DCA+∠ADC=180° (Angle sum property of triangle)

65°+65°+∠ADC=180°

∠ADC=180°-130°=50°

(ii)

AB||CD (given) and BC is transversal

m=20°

In ΔABC (isosceles Δ as AB=BC) 

∠BAC=∠BCA=x

∠BAC+∠BCA+∠ABC=180° (Angle sum property of triangle)

x+x+20°=180°

2x=180°-20°

$x=\frac{160}{2}$

x=80°

In ΔCDE

∠ACD=x+m

∠ACD=80°+20°=100°

n=180°-∠ACD

So, n=180°-100°=80°

In ΔCDE (isosceles)

∠DCE=∠CDE=n=80°

n+n+q=180°

2n+q=180°

2×80°+q=180°

q=180°-160°

q=20°

(iii)

In ΔAEB, AE=EB (isosceles triangle)

∠EAB=∠EBA=20°

∠EAB+∠EBA+∠BEA=180° (Angle sum property of triangle)

20°+20°+∠BEA=180°

∠BEA=180°-40°=140°

or a=140°

b=180°-a (linear pair)

b=180°-140°=40°

In ΔBEF,

∠EBF+∠BEF+∠BFE=180° (Angle sum property of triangle)

60°+40°+∠BFE=180° 

∠BFE=180°-100°=80°

or c=80°

also, EF||DC and (ED and FC are transversal)

∠b=∠m , ∠c=∠n (corresponding angles)

∠m=40° , ∠n=80°

(iv)

In ΔEFH, (isosceles as EF=EH)

∠EFH=∠EHF=35°=y

∠EFH+∠EHF+∠HEF=180° (Angle sum property of triangle)

35°+35°+∠HEF=180°

∠HEF=180°-70°=110°

or z=110°

EF||HG, HF is transversal

y=a (alternate interior angle)

a=35°

also

f=∠EHF (alternate interior angle)

f=35°

In ΔOHG, OG=OH (So, its a isosceles Δ)

∠OHG=∠OGH

∠a=∠OGH

35°=∠OGH

∠HOG+∠OHG+∠OGH=180° (Angle sum property of triangle)

c+35°+35°=180°

c=180°-70°=110°

c+d=180° (linear pair)

d=180°-110°=70°

In ΔOGF,

d+f+e=180° (Angle sum property of triangle)

70°+35°+e=180°

105°+e=180°

e=180°-105°=75°

(v)

AI||BD and AB is transversal

130°+h=180° (same side of transversal)

h=180°-130°=50°

also, 130°=f (corresponding angle)

f=130°

In ΔBDC,

f+g+20°=180° (Angle sum property of triangle)

130°+g+20°=180°

g=180°-150°=30°

(vi)

∠EOD=∠AOB=30° (vertically opposite angle)

In ΔABO ,(isosceles as AO=AB)

∠AOB=∠ABO=30°

∠AOB+∠ABO+∠OAB=180° (Angle sum property of triangle)

30°+30°+∠OAB=180°

∠OAB=180°-60°=120°

or p=120°

OA||BC and OB is transversal

∠AOB=∠OBC

30°=q

or q=30°

∠ABO=∠BOC (alternate interior angle)

30°=r

or r=30°

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Question 12

In the adjoining figure, AE||BC. With the help of the given information find the value of x and y

[Hint: ∠DAE=∠ABC➝ x+19=y-11➝x-y=-30 ....(i)
2x+y+y-11+x+11=180➝3x+2y=180 ...(ii). Solve (i) and (ii)]

Sol :

x+19°=y-11° (corresponding angles)

x=y-11°-19°

x-y=-30°...(i)

In ΔABC,

(2x+y)+(y-11°)+(x+11°)=180° (Angle sum property of triangle)

2x+y+y-11°+x+11°=180°

3x+2y=180°...(ii)

Solving equation (i) and (ii)

$\begin{aligned}3x+2y&=180^{\circ}\\ 2\times [x-y&=-30] \end{aligned}$

or

$\begin{aligned}3x+2y&=180^{\circ}\\2x-2y&=-60\\\hline 5x&=120^{\circ}\end{aligned}$

$x=\frac{120^{\circ}}{5}$

x=24°

Putting value of x in equation (i)

24°-y=-30°

-y=-30°-24°

-y=-54°

y=54°