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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°
y = 135°
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 = 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°
