Selina solutions

Question 1

Q1) Find the unknown angles in the given figures:

Solution:

(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°

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