###
**
Find angle x
**

**
**

In ∆ABC,

AB = AC
*
(Given)
*

Therefore,

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

40° = x

**
x =
**
**
40°
**
**
**

###
**
Find
**
**
angle
**
**
x
**

**
**

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
**

**
**

In ∆XYZ,

XZ = ZY
*
(Given)
*

Therefore,

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

**
x = 50°
**

###
**
Find
**
**
angle x
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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
**

**
**

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°
**