(i) In Fig (i),
x = y (Angles opposite to equal sides)
But x + y + 80° = 180° (Angles of a triangle)
⇒ x + x + 80° = 180°
⇒ 2x = 180° - 80° = 100°
⇒ x = 100°/2 = 50° ∴ y = x = 50°
Hence x = 50°, y = 50°
(ii) In Fig. (ii),
b = 40° (Angles opposite to equal sides)
But a + b + 40° = 180°
(Angles of a triangle)
⇒ a + 40° + 40° = 180°
⇒ a + 80° = 180°
⇒ a = 180° - 80° = 100°
Hence, a = 100° , b = 40°
(iii) In Fig. (iii)
x = y (Angles opposite to equal sides)
But x + y + 90° = 180°
(Angles of a triangle)
⇒ x + x + 90° = 180°
⇒ 2x + 90° = 180°
⇒ 2x = 180° - 90° = 90°
∴ x = 90°/2 = 45° ∴ y = x = 45°
Hence x = 45°, y = 45°
(iv) In Fig. (iv),
a = b (Angles opposite to equal sides)
But a + b + 80° = 180°
⇒ a + a + 80° = 180°
⇒ 2a = 180° - 80° = 100°
⇒ a = 100/2 = 50 ∴ b = a = 50°
x = a + 80°
(Exterior angle of a triangle is equal to sum of its opposite interior angles)
= 50° + 80° = 130°
Hence a = 50°, b = 50° and x = 130°
(v) In Fig. (v),
Let each equal angle of an isosceles triangle
be x,
then, x + x = 86° ⇒ 2x = 86°
⇒ x = 86° /2 = 43°
But p + x = 180° (Linear pair)
p + 43° = 180°
⇒ p = 180° - 43° ⇒ p = 137°
Hence, p = 137°
(vi) In Fig. (vi)
m = 35° (Angles opposite to equal sides)
But m + n + (60° + 35°) = 180°
(Angles of a triangle)
⇒ m + n + 95° = 180°
⇒ 35° + n + 95° = 180°
⇒ n + 130° = 180°
⇒ n = 180° – 130° = 50°
Hence m = 35°, n = 50°
(vii) In Fig. (vii),
x = 60° (Alternate angles)
Let each equal angle of an isosceles triangle
Be a then a + a + x = 180°
(Angles of a triangle)
2a + x = 180° ⇒ 2a + 60° = 180°
⇒ 2a = 180° – 60° = 120°
⇒ a = 120°/2 = 60°
y = x + a = 60° + 60° = 120°
Hence x = 60° and y = 120°