Find angle x

57.jpg

In  ∆ABC,

AB = AC     (Given)

 

Therefore,

∠C = ∠B         (Angles opposite to equal sides are equal)

40° = x

x = 40°   

 

Find angle x

58.jpg

In  ∆PQR,

PQ = QR          (Given)

 

Therefore,

∠R = ∠P           (Angles opposite to equal sides are equal)

45° = ∠P

∠P = 45°

 

Now, by Angle sum property, 

∠P + ∠Q + ∠R = 180°

45° + x + 45° = 180°

x + 90° = 180°

x = 180° − 90°    

x = 90 °

 

Find angle x

59.jpg

In  ∆XYZ,

XZ = ZY         (Given)

 

Therefore,

∠Y = ∠X      (Angles opposite to equal sides are equal)

x = 50°

 

Find angle x

60.jpg

In  ∆ABC,

AB = CA       (Given)

 

Therefore,

∠C = ∠B      (Angles opposite to equal sides are equal)

∠C  = x

 

Now, By angle sum property,

  ∠A + ∠B +∠C = 180°    

   100° + x + x = 180°

   100° + 2x = 180°

  2x = 180° − 100°

  2x = 80°

  x = 80° /2

  x = 40°

 

Find angle x

61.jpg

In  ∆PQR,

QR = PQ      (Given)

 

Therefore,

∠P = ∠R    (Angles opposite to equal sides are equal)

x  = ∠R

∠R = x

 

Now, by angle sum property, 

∠P + ∠Q + ∠R = 180°  

x + 90° + x = 180°

90° + 2x = 180°

2x = 180° − 90°

2x = 90°

x = (90°)/2

x = 45°  

 

Find angle x

62.jpg

In  ∆XYZ,

XY = YZ          (Given)

 

Therefore,

∠Z = ∠X         (Angles opposite to equal sides are equal)

x  = ∠X

∠X = x

 

Now, by angle sum property, 

∠X + ∠Y + ∠Z = 180°  

x + 40° + x = 180°

40° + 2x = 180°

2x = 180° − 40°

2x = 140°

x = (140°)/2

x = 70°  

 

Find angle x

  63.jpg

In  ∆PQR,

PQ = PR      (Given)

 

Therefore,

∠PRQ = ∠Q       (Angles opposite to equal sides are equal)

∠PRQ = x  

 

Now,

∠PRQ + ∠PRS = 180°      (Linear pair)

x + 120° = 180°

x = 180° − 120°

x = 60 °   

 

Find angle x

64.jpg

In  ∆PQR,

PR = PQ       (Given)

 

Therefore,

∠Q = ∠R      (Angles opposite to equal sides are equal)

∠Q = x

 

We know that,

Exterior angle is equal to sum of interior opposite angles

∠SPR = ∠Q + ∠R

110° = x + x

110° = 2x

(110°)/2 = x

55° = x

x = 55° 

 

Find angle x

65.jpg

Here

∠MYN  = ∠ZYX      (Vertically opposite angles)

30°  = ∠ZYX

∠ZYX = 30° 

 

In ∆XYZ,

YZ = ZX           (Given)

 

Therefore,

30°  = ∠ZYX         (Angles opposite to equal sides are equal)

∠X = ∠ZYX

x = 30° 

 

Find angle x and y

66.jpg

In  ∆ABC,

AC = AB                  (Given)

 

Therefore,

∠B = ∠ACB           (Angles opposite to equal sides are equal)

y = ∠ACB  

∠ACB = y

 

Now,

∠ACB + ∠ACD = 180°             (Linear pair)

y + 120° = 180° 

y = 180° − 120°  

y = 60°

 

We know that,

Exterior angle is equal to sum of interior opposite angles.

  ∠ACD = ∠A + ∠B

  120 ° = x + y

  120 ° = x + 60°

  120 ° − 60° = x

  60° = x

  x = 60°   

 

Find angle x and y

68.jpg

In  ∆PQR,

            QR = PR                        (Given)

 

Therefore,

∠QPR = ∠Q                (Angles opposite to equal sides are equal)

∠QPR = x

 

Now,

In Δ PQR

By Angle sum property

∠ QPR + ∠ Q + ∠ R = 180°

x + x + 90° = 180°

2x + 90° = 180°

2x = 180° – 90°

2x = 90°

x = (90°)/2

x = 45°

 

Also,

Exterior angle is sum of interior opposite angles

y = x + ∠ R

 

y = 45° + 90°

 

Find angle x and y

69.jpg

Here,

  ∠JIK = ∠MIN        (Vertically opposite angle)

  ∠JIK = 92°   

 

In ∆IJK,

  IJ = IK                   (Given)

 

Therefore,

  ∠IKJ = ∠IJK        (Angles opposite to equal sides are equal)

  ∠IKJ = x

 

Now, By angle sum property,

 ∠IJK + ∠JKI + ∠KIJ = 180°

  x + x + 92° = 180°

  2x + 92° = 180°

  2x = 180° − 92°

  2x = 88°

  x = (88°)/2

 

∴ x = 45° and y = 135°

x = 44°   

 

Hence,

  IKJ = x

  ∠IKJ = 44°

 

Now,

  ∠IJK + ∠IKL = 180°       (Linear pair)

  44° + y = 180°

  y = 180° − 44° 

  y = 136°

 

x = 44° and y = 136° 

 

  1. Chapter 6 Class 7 Triangle and its Properties
  2. Concept wise
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.